# ML Aggarwal ICSE Solutions Class 10 Math Fifth Chapter Quadratic Equations

### ML Aggarwal ICSE Solutions Class 10 Math 5th Chapter Quadratic Equations

Class 10 Chapter 5 Quadratic Equations

Exercise 5.1

(1) In each of the following, determine whether the given numbers are roots of the given equations or not.

(i) x2 – 5x + 6 = 0; 2, -3

Solution:

Given quadratic equation is x2 – 5x + 6 = 0 —– (1)

To check given numbers are roots of equn (i) or not

We put x = 2 in equn (i)

=> (x)2 – 5x + 6 = (2)2 – 5 (2) + 6

= 4 – 10 + 6

= 10 – 10

= 0

= RHS

Thus, equn (i) satisfies for the value of x = 2.

Hence, x = 2 is the root of quadratic equation (i)

Now, put x = -3 in equn (i)

=> x2 – 5x + 6 = (-3)2 – 5 (3) + 6

= 9 + 15 + 6

= 30

≠ 0

Thus, equn (i) not satisfies for the value of x = -3

Hence, x = -3 not the root of quadratic equn (i)

(ii) 3x2 – 13x – 10 = 0, 5, -2/3

Solution:

Given quadratic equation is 3x2 – 13 – 10 = 0 —— (1)

To check given numbers are roots of equn (i)

Initially, we put x = 5 in equn (i)

=> 3x2 – 13x – 10 = 3 (5)2 – 13 (5) – 10

= 3 (25) – 65 – 10

= 75 – 75

= 0

Thus, equn (ii) satisfies for the value of x = 5.

Hence, x = 5 is the root of equation (i)

Now, we will put x = -2/3 in equn (i)

3x2 – 13x – 10 = 3 (-2/3)2 – 13 (-2/3) – 10

= 3 (4/9) + 26/3 – 10

= 12/9 + 26/3 – 10

= 4/3 + 26/3 – 10

= 30/3 – 10

= 10 – 10

= 0

Thus, equn (i) satisfies for the value of x = -2/3

Hence, x = -2/3 is the root of equn (i).

(2) In each of the following, determine whether the given numbers are solutions of equation or not.

(i) x2 – 3√3 x + 6 = 0; √3, -2√3

(ii) x2 – √2 x – 4 = 0, x = – √2, 2√2

Solution:

x2 – 3√3 x + 6 = 0 —– (i)

To check given numbers are roots of equation (i),

We put x = √3 in equn (i)

=> x2 – 3√3 x + 6 = (√3)2 – 3√3 (√3) + 6

= 3 – 9 + 6

= 9 – 9

= 0

Thus, equn (i) satisfies for the value of x = √3.

Hence, x = √3 is the root of equation (i)

Now, put x = -2√3 in equn (i)

=> x2 – 3√3 + 6 = (-2√3)2 – 3√3 (-2√3) + 6

= 12 + 18 + 6

= 36

≠ 0

Thus, equn (i) not satisfies for the value of x = -2√3

Hence, x = -2√3 is not the root of equn (i)

x2 – √2 x – 4 = 0, x = -√2, 2√2 —– (i)

To check given numbers are roots of equn (i)

We put x = – √2 in equn (i)

=> x2 – √2x – 4 = — (-√2)2 – √2 (-√2) – 4

= 2 + 2 – 4

= 4 – 4

= 0

Thus, equn (i) satisfies for the value of x = -√2

Hence, x = – √2 is the root of equn (i)

Now, put x = 2√2 in equn (i)

=> x2 – √2x – 4 = (2√2)2 – √2 (2√2) – 4

= 8 – 4 – 4

= 8 – 8

= 0

Thus, equn (i) satisfies the value of x = 2√2

Hence, x = 2√2 is the root of equn (i).

(3) (i) If -1/2 is a solution of the equation 3x2 + 2kx – 3 = 0, find the value of K.

Solution:

Given quadratic equation is 3x2 + 2kx – 3 = 0 —– (i)

Also, x = -1/2 is the solution of equation (i)

To find K, we put x = – 1/2 in equn (i)

=> 3x2 + 2kx – 3 = 0

3 (-1/2)2 + 2k (-1/2) – 3 = 0

3/4 – k – 3 = 0

3/4 = k+3

3/4 – 3 = k

k = -9/4 is the required value of k.

(ii) If 2/3 is a solution of the equation 7x2 + kx – 3 = 0, find the value of k.

Solution:

7x2 + kx – 3 = 0 —– (i)

Also, x = 2/3 is the solution of equn (i)

To find k, we put x = 2/3 in equn (i)

=> 7x2 + kx – 3 = 0

7x2 + kx – 3 = 0

7 (2/3)2 + k (2/3) – 3 = 0

28/9 + 2/3 k – 3 = 0

2/3 k = 3 – 28/9

2/3 k = -1/9

2k = -1/3

k = -1/6 is the required value of k.

(4) (i) If √2 is a root of the equation kx2 + √2x – 4 = 0, find the value of k.

Solution:

Given quadratic equation is kx2 + √2x – 4 = 0 —– (i)

Given that, x = √2, is the solution of (i)

We put x = √2 in equn (i)

=> kx2 + √2 x – 4 = 0

k (√2)2 + √2 (√2) – 4 = 0

2k + 2 – 4 = 0

2k – 2 = 0

K = 1 is the required value of k.

(ii) If ‘a’ is a root of the equation x2 – (a + b) x + k = 0, find the value of k.

Solution:

x2 – (a + b) x + k = 0 —– (i)

x = a is the solution of equation (i)

To find k, put x = a in equn (i)

x2 – (a + b) x + k = 0

a2 – (a + b) a + k = 0

k = – a2 + (a + b) a

k = -a2 + (a + b) a

k = -a2 + a2 + ab

k = ab is the required value of k.

(5) If 2/3 and -3 are the roots of the equation px2 + 7x + q = 0, find the values of P and q.

Solution:

px2 + 7x q = 0 —- (i) Exercise – 5.2

(1) (i) x2 – 3x – 10 = 0 Solve by the method of factorization.

Solution:

Given quadratic equation is x2 – 3x – 10 = 0

x2 – 3x – 10 = 0

x2 – 5x + 2x – 10 = 0

x (x – 5) + 2 (x – 5) = 0

(x + 2) (x – 5) = 0

x = -2 or x = 5

(ii) x (2x + 5) = 3

Solution:

Given equation is

x (2x + 5) = 3

2x2 + 5x – 3 = 0

2x2 + 6x – x – 3 = 0

2x (x + 3) – 1 (x + 3) = 0

(x + 3) (2x – 1) = 0

x = -3 or x = 1/2

(2) (i) 3x2 – 5x – 12 = 0

Solution:

Given equation is

3x2 – 5x – 12 = 0

3x2 – 9x + 4x – 12 = 0

3x (x – 3) + 4 (x – 3) = 0

(x – 3) (3x + 4) = 0

x – 3 = 0 or 3x + 4 = 0

x = 3 or x = -4/3

(ii) 21x2 – 8x – 4 = 0

Solution:

21x2 – 8x – 4 = 0

21x2 – 14x + 6x – 4 = 0

7x (3x – 2) + 2 (3x – 2) = 0

(3x – 2) (7x + 2) = 0

3x – 2 = 0 or 7x + 2 = 0

x = 2/3 or x = -2/7

(3) (i) 3x2 = x + 4

Solution:

Given equn is 3x2 – x – 4 = 0

3x2 – 4x + 3x – 4 = 0

x (3x – 4) + 1 (3x – 4) = 0

(3x – 4) (x + 1) = 0

3x – 4 = 0 or x + 1 = 0

x = 4/3 or x = -1

(ii) x (6x – 1) = 35

Solution:

Given equn is 6x2 – x – 35 = 0

6x2 – x – 35 = 0

6x2 – 15x + 14x – 35 = 0

3x (2x – 5) + 7 (2x – 5) = 0

(2x – 5) (3x + 7) = 0

x = 5/2 or x = -7/3

(4) (i) 6p2 + 11p – 10 = 0

Solution:

Given equn is 6p2 + 11p – 10 = 0

6p2 + 15p – 4p – 10 = 0

3p (2p + 5) – 2 (2p + 5) = 0

(2p + 5) (3p – 2) = 0

2p + 5 = 0 or 3p – 2 = 0

2p = -5 or 3p = 2

p = -5/2 or p = 2/3

(ii) 2/3 x2 – 1/3 x – 1 = 0

Solution:

2/3 x2 – 1/3 x – 1 = 0

2x2 – x – 3 = 0

2x2 – 3x + 2x – 3 = 0

x (2x – 3) + 1 (2x – 3) = 0

(2x – 3) (x + 1) = 0

2x – 3 = 0 or x + 1 = 0

x = 3/2 or x = -1

(5) (i) 3 (x – 2)2 = 147

Solution:

Given equation is

3 (x2 – 4x + 4) = 147

3x2 – 12x + 12 = 147

3x2 – 12x + 12 – 147 = 0

3x2 – 12x – 135 = 0

x2 – 4x – 45 = 0

x2 – 9x + 5x – 45 = 0

x (x – 9) + 5 (x – 9) = 0

(x – 9) = 0 or (x + 5) = 0

x = 9 or x = -5

∵ (a + b)2 = a2 + 2ab + b2

(a – b)2 = a2 – 2ab + b2

(ii) 7 (3x – 5)2 = 28

Solution:

Given equation is

1/7 (3x – 5)2 = 28

(3x – 5)2 = 4

9x2 – 30x + 25 = 4

9x2 – 30x + 25 – 4 = 0

9x2 – 30x + 21 = 0

3x2 – 7x – 3x + 7 = 0

x (3x – 7) – 1 (3x – 7) = 0

(3x – 7) (x – 1) = 0

3x – 7 = 0 or x – 1 = 0

x = 7/3 or x = 1

(6) x2 – 4x – 12 = 0, XEN

Solution:

x2 – 4x – 12 = 0 —– (i)

x2 – 6x + 2x – 12 = 0

x (x – 6) + 2 (x – 6) = 0

(x – 6) (x + 2) = 0

x = 6 or x = -2

But XEN

Hence, x = 6 is the only solution of equn (i).

(7) 2x2 – 9x + 10 = 0

(i) XEN

(ii) XES

Solution:

Given quadratic equn is 2x2 – 9x + 10 = 0 —– (i)

2x2 – 5x – 4x + 10 = 0

x (2x – 5) – 2 (2x – 5) = 0

(2x – 5) (x – 2) = 0

x = 5/2 or x = 2

(i) When XEN

x = 2 is the only one solution of equation (i)

(ii) When XEQ then x = 2, 5/2 is the solution of equn (i)

(8) (i) a2 x2 + 2ax + 1 = 0, a ≠ 0

Solution:

a2 x2 + 2ax + 1 = 0 —- (i)

a2 x2 + ax + ax + 1 = 0

ax (ax + 1) + 1 (ax + 1) =

(ax + 1)2 = 0

ax = -1

x = -1/a

Thus, x = -1/a, -1/a is the required soln.

(ii) x2 – (p+q) x + pq = 0

Solution:

x2 – (p+q) x + pq = 0 —- (i)

x2 – px – qx + pq = 0

x (x – p) – q (x – p) = 0

(x – p) (x – q) = 0

(x – p) = 0 or (x – q) = 0

x = p or x = 1 is the required solution of equn (i)

(9) a2 x2 + (a2 + b2) x + b2 = 0, a ≠ 0

Solution:

a2 x2 + (a2 + b2) x + b2 = 0, a ≠ 0

a2 x2 + a2 x + b2 x + b2 = 0

a2 x (x + 1) + b2 (x + 1) = 0

(x + 1) (a2x + b2) = 0

(x + 1) = 0 or a2x + b2 =

x = -1 or a2x = -b2

x = -b2/a2

This is the required solution of given quadratic equn.

(10) (i) √3x2 + 10x + 7√3 = 0

Solution:

√3x2 + 10x + 7√3 = 0 —- (i)

√3x2 + 7x + 3x + 7√3 = 0

√3x2 + 3x + 7x + 7√3 = 0

√3x (x + √3) + 7 (x + √3) = 0

(x + √3) (-√3x + 7) = 0

x + √3 = 0 or √3x + 7 = 0

x = – √3 or x = 7/√3

This is the required solution of given equation.

(ii) 4√3x2 + 5x – 2√3 =

Solution:

4√3 x2 + 5x – 2√3 = 0 — (i)

4√3x2 + 8x – 3x – 2√3 = 0

4x (√3x + 2) – √3 (√3x + 2) = 0

(√3x + 2) (4x – √3) = 0

√3x + 2 = 0 or 4x – √3 = 0

x = -2/√3 or x = √3/4

This is the required solution of given equation.

(11) (i) x2 – (1 + √2) x + √2 = 0

Solution:

Given equation is

x2 – (1 + √2) x + √2 = 0

x2 – x – √2x + √2 = 0

x (x – 1) – √2 (x – 1) = 0

(x – 1) (x – √2) = 0

x – 1 = 0 or x – √2 = 0

x = 1 or x = √2

This is the required solution of given equation.

(ii) x + 1/x = 2 1/20

Solution:

Given equation is

x + 1/x = 2 1/20

x2+1/x = 41/20

20x2 + 20 = 41x

20 (x2 + 1) – 41x = 0

20x2 – 41x + 20 = 0

20x2 – 25x – 16x + 20 = 0

5x (4x – 5) – 4 (4x – 5) = 0

(4x – 5) (5x – 4) = 0

x = 5/4 or x = 4/5

This is the required solution of given equation.

(12) (i) 2/x2 – 5/x + 2 = 0, x ≠ 0

Solution:

Given quadratic equn is 2/x2 – 5/x + 2 = 0, x≠0

2 – 5x + 2x2 = 0 ∵ multiply by x2

2x2 – 5x + 2 = 0

2x2 – 4x – x + 2 = 0

2x (x – 2) – 1 (x – 2) = 0

(x – 2) (2x – 1) = 0

x – 2 = 0 or 2x – 1 = 0

x = 2 or x = 1/2

This is the required solution of given equn.

(13) (i) 3x – 8/x = 2

Solution:

Given equation is

3x – 8/x = 2

(3x2 – 8)/x = 2

3x2 – 8 = 2x

3x2 – 2x – 8 = 0

3x2 – 6x + 4x – 8 = 0

3x (x – 2) + 4 (x – 2) = 0

(x – 2) (3x + 4) = 0

x – 2 = 0 or 3x + 4 = 0

x = 2 or x = -4/3

This is the required solution of given equation.

(ii) x+2/x+3 = 2x-3/3x-7

Solution:

Given equation is x+2/x+3 = 2x-3/3x-7

(x+2) (3x – 7) = (2x – 3) (x + 3)

(3x2 – 7x + 6x – 14) = (2x2 + 6x – 3x – 9)

3x2 – 2x2 – x – 3x – 14 + 9 = 0

x2 – 4x – 5 = 0

x2 – 5x + x – 5 = 0

x (x-5) + 1 (x – 5) = 0

(x – 5) (x + 1) = 0

x – 5 = 0 or x + 1 = 0

x = 5 or x = -1

This is the required solution of given equation.

(14) (i) 8/(x+3) – 3/(2-x) = 2

Solution:

Given equation is

8/(x+3) – 3/(2-x) = 2

8 (2-x)-3(x+3)/(x+3)(2-x) = 2

(16 – 8x) – 3x – 9 = 2 (x + 3) (2 – x)

7 – 11x = 2 (2x – x2 + 6 – 3x)

7 – 11x = (4x – 2x2 + 12 – 6x)

7 – 11 = -2x – 2x2 + 12

2x2 – 9x – 5 = 0

2x2 – 10x + x – 5 = 0

2x (x – 5) + 1 (x – 5) = 0

(x – 5) (2x + 1) = 0

x – 5 = 0 or 2x + 1 = 0

x = 5 or = x = -1/2

(ii)  x/x-1 + x-1/x = 2 1/2

Solution:

Given equation is

x/x-1 + x-1/x = 5/2

x2+(x-1)2/x(x-1) = 5/2

2 (x2 + x2 – 2x + 1) = 5 (x2 – x)

2 (2x2 – 2x + 1) = 5x2 – 5x

4x2 – 4x + 2 = 5x2 – 5x

x2 – x – 2 = 0

x2 – 2x + x – 2 = 0

x (x – 2) + 1 (x – 2) = 0

(x – 2) (x + 1) = 0

x – 2 = 0 or x + 1 = 0

x = 2 or x = -1

(15) (i) x+1/x-1 + x-2/x+2 = 3

Solution:

Given equation is 2x2 + 4 = 3x2 + 3x – 6

x2 + 3x – 10 = 0

x2 + 5x – 2x – 10 = 0

x (x + 5) – 2 (x + 5) = 0

(x + 5) (x – 2) = 0

x + 5 = 0 or x – 2 = 0

x = -5 or x = 2

(ii) 1/x-3 – 1/x+5 = 1/6

Solution:

Given equation is

1/x-3 – 1/x+5 = 1/6 48 = x2 + 2x – 15

x2 + 2x – 63 = 0

x2 + 9x – 7x – 63 = 0

x (x + 9) – 7 (x + 9) = 0

(x + 9) (x – 7) = 0

x + 9 = 0 or x – 7 = 0

x = -9 or x = 7

(16) a/ax-1 + b/(bx-1) = (a+b), a + b ≠ 0, ab≠0

Solution:

Given equation is a/(ax-1) + b/(bx-1) = (a + b)

a (bx-1)+b(ax-1)/(ax-1)(bx-1) = a + b

abx – a + abx – b/(abx2 – ax – bx + 1) = (a+b)

2abx – a – b = a2bx2 – a2x – abx + a + ab2x2 – abx – b2x + b

a2bx2 + ab2x2 – a2x – abx – b2x + a + b + a + b – 2abx = 0

(a2b + ab2) x2 – (a2 + 2ab + b2) x + 2a + 2b – 2abx = 0

(a2b + ab2) x2 – (a2 + 4ab + b2) x + 2 (a + b) = 0 (17) 1/x+6 + 1/x-10 = 3/x-4

Solution:

Given equation is 1/x+6 + 1/x-10 = 3/x-4

x-10+x+6/(x+6)(x-10) = 3/x-4

2x-4/(x2-10x+6x-60) = 3/x-4

(2x-4) (x – 4) = 3 (x2 – 4x – 60)

(2x2 – 8x – 4x + 16) = 3x2 – 12x – 180

x2 – 180 – 16 = 0

x2 – 196 = 0

x2 = 196

x = √196

x = ±14

This is the required value of x.

(18) (i) √3x+4 = x

Solution:

√3x+4 = x

Squaring on both sides,

(3x + 4) = x2

x2 – 4x + x – 4 = 0

x (x – 4) + 1 (x – 4) = 0

(x – 4) (x + 1) = 0

x – 4 = 0 or x + 1 = 0

x = 4 or x = -1

(ii) √x(x-7) = 3√2

Solution:

Given equation is

√x(x-7) = 3√2

Squaring on both sides,

x (x-7) = 18

x2 – 7x = 18

x2 – 7x – 18 = 0

x2 – 9x + 2x – 18 =0

x (x – 9) + 2 (x – 9) = 0

(x – 9) (x + 2) = 0

x – 9 = 0 or x + 2 = 0

x = 9 or x = -2

(19) Use the substitution y = 3x + 1 to solve for x:

5 (3x + 1)2 + 6 (3x + 1) – 8 = 0

Solution:

Given that, 5 (3x + 1)2 + 6 (3x + 1) – 8 = 0 —- (i)

Put y = 3x + 1 in (i)

5y2 + 6y – 8 = 0

5y2 + 10y – 4y – 8 = 0

5y (y + 2) – 4 (y + 2) = 0

(y + 2) (5y – 4) = 0

y + 2 = 0 or 5y – 4 = 0

y = 2 or y = 4/5

But y = 3x + 1

=> 3x + 1 = -2 or 3x + 1 = 4/5

3x = -3 or 3x = 4/5 – 1 = -1/5

x = -1 or x = -1/15

This are the required values of x for equation (i).

(20) Find the values of x, if p + 1 = 0 and x2 + px – 6 = 0

Solution:

Given that, p + 1 = 0

P = -1 put in given equn

=> x2 + px – 6 = 0

x2 – x – 6 = 0

x2 – 3x + 2x – 6 = 0

x (x – 3) + 2 (x – 3) = 0

(x – 3) (x + 2) = 0

x – 3 = 0 or x + 2 = 0

x = 3 or x = -2

This are the required values of x for given equation.

(21) Find the values of x if P + 7 = 0, q = 12 & x2 + px + q = 0

Solution:

Given equation is x2 + px + q = 0 —– (i)

Also, p + 7 = 0 and q = 12

=> p = -7

By putting the values of p & q in equn (i)

=> x2 – 7x + 12 = 0

x2 – 4x – 3x + 12 = 0

x (x – 4) – (x – 4) = 0

(x – 4) (x – 3) = 0

x – 4 = 0 or x – 3 = 0

x = 4 or x = 3

This are the required values of x for given equn

(22) If x = 3 is a solution of equn (k + 2) x2 – kx + 6 = 0

Find the value of k & find the other root of equation also.

Solution:

Given equation is (k + 2) x2 – kx + 6 = 0 —- (i)

x = 3 is the solution of equn (i)

Put x = 3 in equn (i) => (k + 2) (3)2 – k (3) + 6 = 0

9 (k + 2) – 3k + 6 = 0

9k + 18 – 3k + 6 = 0

6k + 24 = 0

6k = -24

k = -4

Then, put k = -4 in equn (i) => (-4+2) x2 + 4x + 6 = 0

-2x2 + 4x + 6 = 0

2x2 – 4x + 6 = 0

x2 – 2x – 3 = 0

x2 – 3x + x – 3 = 0

x (x – 3) + 1 (x – 3) = 0

(x – 3) (x + 1) = 0

x – 3 = 0 or x + 1 = 0

x = 3 or x = -1

Thus, the another root of equn (i) is found to be x = -1

(23) If x = p is a solution of equn x (2x + 5) = 3, then find the value of p.

Solution:

Given, equn is x (2x + 5) = 3

2x2 + 5x – 3 = 0 —– (i)

2x2 + 6x – x – 3 = 0

2x (x + 3) – 1 (x + 3) = 0

(x + 3) (2x – 1) = 0

x + 3 = 0 or 2x – 1 = 0

x = -3 or x = 1/2

But, given that x = p

Thus, the required values of p for given equation (i) are found to be p = -3 and p = 1/2.

Exercise – 5.3

(i) 2x2 – 7x + 6 = 0

Solution:

2x2 – 7x + 6 = 0 —— (i)

2x2 – 3x – 4x + 6 = 0

2x2 – 4x – 3x + 6 = 0

2x (x – 2) – 3 (x – 2) = 0

(x – 2) (2x – 3) = 0

x – 2 = 0 or 2x – 3 = 0

x = 2 or x = 3/2

This are the required values of x for the given equation (i).

(ii) 2x2 – 6x + 3 = 0

Solution:

Given equation is

2x2 – 6x + 3 = 0 —– (i)

Comparing equn (i) with ax2 + bx + c = 0

We get, a = 2, b = -6, c = 3 This are the required values of x for give equation (i).

(2) (i) 25x2 + 30x + 7 = 0

Solution:

Given equn is = 25x2 + 30x + 7 = 0 —- (i)

On comparing equn (i) with ax2 + bx + c = 0

We get, a = 25, b = +30, c = 7 This are the required values of x for given equation (i)

(3) (i) 2x2 + √5x – 5 = 0

Solution:

Given equation is

2x2 + √5x – 5 = —- (i)

On comparing equn (i) with

ax2 + bx + c = 0

=> a = 2, b = √5, c = 5

b2 – 4ac = 5 – 4 (2) (-5)

= 5 + 40

b2 – 4ac = 45

Now, This are the required roots of given equation (i).

(ii) √3x2 + 10x – 8√3 = 0

Solution:

Given equation is

√3x2 + 10x – 8√3 = 0 —- (i)

On comparing equn (i) with ax2 + bx + c = 0

We get a = √3, b = 10, c = 8√3

b2 – 4ac = 100 – 4 (√3) (-8√3)

= 100 + 32 × 3

= 100 + 96

b2 – 4ac = 196

Now, This are the required roots of given equation (i).

(5) (i) 4x2 – 4ax + (a2 – b2) = 0

Solution:

Given equation is 4x2 – 4ax + (a2 – b2) 0 —- (i)

On comparing equn with ax2 + bx + c = 0

We get a = 4, b = -4a, c = a2 – b2

Now, b2 – 4ac = 16a2 – 4 (4) (a2-b2)

= 16a2 – 16a2 + 16b2

b2 – 4ac = 16b2

√b2 – 4ac = 4b

Then, x = -b±√b2 – 4ac/2a

x = 4a±4b/8 = a±b/2

x = a+b/2 or x = a-b/2

This are the required values of x for given equation (i)

(6) (i) x – 1/x = 3, x≠0

Solution:

Given equation is

x – 1/x = 3 — (i)

x2-1/x = 3

x2 – 1 = 3x

x2 – 3x – 1 = 0

Here, a = 1, b = -3, c = -1

x = -b±√b2-4ac/2a

x = 3±√9+4/2

x = 3±√13/2

x = 3±√13/2 or x = 3-√13/2

This are the required values of x for given equn (i)

(ii) 1/x + 1/x-2 = 3, x≠0, 2

Solution:

Given equation is

1/x + 1/x-2 = 3 —- (i)

(x – 2) + x/x (x-2) = 3

2x – 2/x2 – 2x = 3

2x – 2 = 3x2 – 6x

3x2 – 6x – 2x + 2 = 0

3x2 – 8x + 2 = 0

Here, a = 3, b = -8, c = +2 This are the required values of x for given equn (i)

(i) x2 – 5x – 10 = 0

Solution:

Given equation is x2 – 5x – 10 = 0 —- (i)

On comparing equn(i) with ax2 + bx + c = 0

We get, a = 1, b = -5, c = -10

Now, b2 – 4ac = (-5) × 4 (1) (-10) = 25 + 40 = 65 There are the required values of x for given Equn (i)

(9) (i) 4x2 – 5x – 3 = 0

Solution:

Given equation is 4x2 – 5x – 3 = 0 —- (i)

On comparing equn(i) with ax2 + bx + c = 0

We get, a = 4, b = -5, c = -3

Then, b2 – 4ac = (-5)2 – 4 (4) (-3) = 25 + 48 = 73

Now, x = -b±√b2-4ac/2a

x = 5±√73/8

x = 5+√73/8 or x = 5-√73/6

x = 5+8.54/8 or x = 5-8.54/8

x = 1.69 or x = -0.44

These are the required values of x for given equn(i).

(10) (i) x2 – 4x – 8 = 0

Solution:

Given equn is x2 – 4x – 8 = 0 —— (i)

On comparing equn (i) with ax2 + bx + c = 0

a = 1, b = -4, c = -8

b2 – 4ac = (-4)2 – 4 (1) (-8)

= 16 + 32

= 48

Now,

x = -b±√b2-4ac/2a

x = 4±√16+32/2 = 4±√48/2

x = 4±4√3/2 = 2±2√3

x = 5.465 or x = -1.465

(ii) x – 18/x = 6

Solution:

Given equn is

x – 18/x = 6 —- (i)

On comparing equn (i)

With ax2 + bx + c = 0

x2 – 18/x = 6

x2 – 18 = 6x

x2 – 6x – 18 = 0 —– (i)

a = 1, b = -6, 0 = -18

b2 – 4ac = 36 – 4 (1) (-18)

= 36 + 72

b2 – 4ac = 108

Now, (11) Solve the equn 5x2 – 3x – 4 = 0 & give your answer correct to 3 significant figures.

Solution:

Given equn is 5x2 – 3x – 4 = 0 —- (i)

On comparing equn (i) with ax2 – bx + c = 0

a = 5, b = -3, c = -4

b2 – 4ac = 9 – 4 (5) (-4)

= 9 + 80 = 89

Now, x = -b±√b2-4ac/2a = 3±√89/10

x = 3+√89/10 or x = 3-√89/10

x = 1.24 or x = -0.643

These are the required of x for given equn (i).

Exercise 5.4

(1) Find the discriminant of the following equations & hence find the nature of the roots.

(i) 3x2 – 5x – 2 = 0

Solution:

Here, given equn is

3x2 – 5x – 2 = 0 —– (i)

On comparing equn (i) with

ax2 + bx + c = 0

=> a = 3, b = -5, c = -2

D = b2 – 4ac

= (-5)2 – 4(3) (-2)

= 25 + 24

D = 49

D > 0

=> The roots are real & district

(ii) 2x2 – 3x + 5 = 0

Solution:

Here, given equation is

2x2 – 3x + 5 = 0 —– (i)

On comparing equn (i) with

ax2 + bx + c = 0

=> a = 2, b = -3, c =

D = b2 – 4ac

= (-3)2 – 4 (2) (5)

= 9 – 40

D = -31

D < 0

The roots are imaginary or not real.

(2) Discuss the nature of the roots of the following quadratic equations:

(i) 3x2 – 4√3x + 4 = 0

Solution:

Given equn is 3x2 – 4√3x + 4 = 0 —— (i)

On comparing equn (i) with

ax2 + bx + c = 0

=> a = 3, b = -4√3, c = 4

D = b2 – 4ac = (-4√3)2 – 4 (3) (4)

= 48 – 48

D = 0

Thus, the roots are real & equal.

(ii) x2 – 1/2x + 4 = 0

Solution:

Given equation is

x2 – 1/2 x + 4 = 0 —– (i)

On comparing equn (i) with

ax2 + bx + c = 0

=> a = 1, b = -1/2, c = 4

D = b2 – 4ac

= 1/4 – 4 (i) (4) = 1/4 – 16

D = (1 – 64)/4 = -63/4

D = – 63/4

D < 0

Thus, the roots are not real or imaginary.

(3) Find the nature of the roots of the following of quadratic equations.

(i) x2 – 1/2 x – 1/2 = 0

Solution:

Given quadratic equn is x2 – 1/2x – 1/2 = 0 —– (1)

2x2 – x – 1 = 0 —– (2)

On comparing equn (2) with ax2 + bx + c = 0

a = 2, b = -1, c = -1

D = b2 – 4ac = (-1)2 – 4 (2) (-1)

D = 1 + 8 = 9

D = 9

D > 0

Thus, the roots are real and distinct.

(ii) x2 – 2√3x – 1 = 0, if real roots exist, find them.

Solution

Given equation is x2 – 2√3x – 1 = 0 —- (1)

On comparing equn(1) with ax2 + bx + c = 0

=> a = 1, b = -2√3, c = -1

D = b2 – 4ac = (-2√3)2 – 4 (1) (-1)

= 12 + 4

D = 16

D > 0

Thus, the roots are real & distinct.

(4) Without solving the quadratic equation, find the value of ‘p’ for which the given equations have real & equal roots:

(i) Px2 – 4x + 3 = 0

Solution:

Given equation is px2 – 4x + 3 = 0 —- (i)

Here, a = p, b = -4, c = 3

Given that equn(1) has real & equal roots => D = 0

D = b2 – 4ac

0 = (-4)2 – 4 (p) (3)

0 = 16 – 12p

12p = 16

P = 4/3

16/12 = 4/3

(ii) x2 + (p – 3) x + p = 0

Solution:

Here, a = 1, b = (p – 3) c = p

Given that, roots are real & equal

=> D = 0

D = b2 – 4ac

0 = P2 – 6p + 9 – 4p

p2 – 10p + 9 = 0

P2 – 10p + 9 = 0

P2 – 9p – p + 9 = 0

P (p – 9) + 1 (p – 9) = 0

(p – 9) (p+1) = 0

P – 9 = 0 or P + 1 = 0

P = 9 or P = +1

(5) (i) Find the value of K for which following quadratic equation has equal roots:

x2 + 4kx + (k2 – k + 2) = 0

Solution:

x2 + 4kx + (k2 – k + 2) = 0 —– (1)

On comparing equn (1) with ax2 + bx + c = 0

=> a = 1, b = 4k, c = k2 – k + 2

Given, that roots of equn (1) are equal => D = 0

D = b2 – 4ac = (4k)2 – 4 (1) (k2 – k + 2)

= 16k2 – 4 (k2 – k + 2)

= 16k2 – 4k2 + 4k – 8

0 = 12k2 + 4k – 8

=> 3k2 + k – 2 = 0

3k2 + 3k – 2k – 2 = 0

3k2 + 3k – 2k – 2 = 0

3k (k + 1) – 2 (k + 1) = 0

(k + 1) (3k – 2) = 0

(k + 1) = 0 or 3k – 2 = 0

k = -1 or k = 2/3

These are the required values of K for given equation.

(6) Find the values of M for which the following quadratic equation has real & equal roots.

(3m + 1) x2 + 2 (m + 1) x + m = 0

Solution:

Given quadratic equation is (3m + 1)x2 + 2 (m + 1) x + m = 0

On comparing equn (1) with ax2 + bx + c = 0

=> a = (3m + 1), b = 2 (m + 1), c = m

Then, D = b2 – 4ac

= 22 (m + 1)2 – 4 (3m + 1) (m)

= 4 (m2 + 2m + 1) – (12m2 + 4m)

= 4m2 + 8m + 4 – 12m2 – 4

D = -8m2 + 4m + 4

But, given that equation (1) has real & equal roots

=> D = 0 => -8m2 + 4m + 4 = 0

2m2 – m – 1 = 0

2m2 – 2m + m – 1 = 0

2m (m – 1) + 1 (m – 1) = 0

(m – 1) (2m + 1) = 0

m = 1 or m = -1/2

These are the required values of m for given equation (1)

(7) Find the values of K for which following quadratic equation has equal roots: 9x2 + kx + 1 = 0

Solution:

Given quadratic equation is 9x2 + kx + 1 = 0 — (1)

On comparing equn with ax2 + bx + c = 0

=> a = 9, b = k, c = 1

Then, D = b2 – 4ac = K2 – 36

But, given that roots are equal => D = 0

K2 – 36 = 0

K2 = 36

K = ±6

(8) Find the values of P for which the quadratic equation (2p + 1) x2 – (7p + 2) x + (7p – 3) = 0 has equal roots. Also find these roots

Solution:

(2p + 1) x2 – (7p + 2) x + (7p – 3) = 0 —– (1)

On comparing equn (1) with ax2 + bx + c = 0

=> a = (2p + 1), b = – (7p + 2), c = 7p – 3

D = b2 – 4ac

= (7p + 2)2 – 4 (2p + 1) (7p – 3)

= 49p2 + 28p + 4 – 4 (14p2 – 6p + 7p – 3)

= 49p2 + 28p + 4 – 56p2 + 24p – 28p + 12

= -7p2 + 28p – 4p + 16

D = -7p2 + 24p + 16

But, given that, the roots are equal => D = 0

=> -7p2 + 24p + 16 = 0

7p2 – 24p – 16 = 0

7p2 – 28p + 4p – 16 = 0

7p (p – 4) + 4 (p – 4) = 0

(p – 4) (7p + 4) = 0

p = 4 or p = -4/7

These are the required values of P for given equation (1)

P = 4

=> 9x2 – 30x + 25 = 0

9x2 – 15x – 15x + 25 = 0

3x (3x – 5) – 5 (3x – 5) = 0

(3x – 5) (3x – 5) = 0

3x – 5 = 0 or 3x – 5 = 0

x = 5/3 or x = 5/3

These are the roots of given equation (1)

(9) Find the values of P for which the equation 2x3 + 3x + p = 0 has real roots.

Solution:

Given quadratic equation is 2x2 + 3x + p = 0 —– (1)

On comparing (1) with ax2 + bx + c = 0

∴ a = 2, b = 3, c = p

Then, D = b2 – 4ac

= 9 – 4 (2) p

D = 9 – 8 p (But given that roots of equn (1) are equal.

=> D = 0

9 – 8p = 0

8p = +9

P = 9/8 is the required value of P when roots are real & equal.

But, when roots are real only => D≥0

9 – 8p ≥ 0

9 ≥ 8p

P ≤ 9/8 is the required value of p.

(10) Find the least positive value of K for which the equn x2 + kx + 4 = 0 has real roots.

Solution:

Given quadratic equation is x2 + kx + 4 = 0 —- (1)

On comparing equn (1) with ax2 + bx + c = 0

∴ a = 1, b = k, c = 4

D = b2 – 4ac = K2 – 16

But, given that roots are real = D≥0

= k2 – 16 ≥ 0

k2 ≥ 16

k ≥ 4

k = 4 is the required least positive value of for equation (1)

(11) Find the values of P for which the equation 3x2 = Px + 5 = 0 has real roots.

Solution:

Given quadratic equation is 3x2 – Px + 5 = 0 —- (1)

On comparing equn (1) with ax2 + bx + c = 0

=> a = 3, b = -p, c = 5

D = b2 – 4ac = p2 – 4 (3) (5)

D = p2 – 60

But, given that roots are real => D≥0

= p2 – 60 ≥ 0

p2 ≥ 60

p ≥ ± √60

P ≥ ± 2√15

p ≥ + 2√15 or p ≤ -2√15

These are the required values of P for given equn (1)

Exercise 5.5

(1) Find the two consecutive natural numbers such that the sum of their squares is 61.

Solution:

Let us consider ‘x’ is the first natural no.

Then (x+1) is next consecutive natural no.

From given condition, we can write

x2 + (x + 1)2 = 61

x2 + x2 + 2x + 1 = 61

2x2 + 2x + 1 = 61

2x2 + 2x – 60 = 0

x2 + x – 30 = 0

x2 + 6x – 5x – 30 = 0

x (x + 6) – 5 (x + 6) = 0

(x + 6) (x – 5) = 0

x = -6 or x = 5 then x + 1 = 5 + 1 = 6

Thus, the required consecutive natural no. are 5 & 6.

(2) (i) If the product of two consecutive even integers is 224, find the integers.

Solution:

Let us consider first even integer be ‘2x’

Then next consecutive even integer be (2x + 2)

From given condition,

2x (2x + 2) = 224

4x2 + 4x = 224

4x2 + 4x – 224 = 0

x2 + x – 56 = 0

x2 + 8x – 7x – 56 = 0

x (x + 8) – 7 (x + 8) = 0

(x + 8) (x – 7) = 0

x + 8 = 0 or x – 7 = 0

x = – 8 or x = 7

Then 2x = 14 & 2x + 2 = 14 + 2 = 16

Thus, the required two consecutive even integers are 14 & 16 respectively.

(ii) Find the two consecutive odd integer such that the sum of their squares is 394.

Solution:

Let us consider the first odd integer be (2x + 1)

Then next consecutive odd integer be (2x + 3)

From given condition,

(2x + 1)2 + (2x + 3)2 = 394

4x2 + 4x + 1 + 4x2 + 12x + 9 = 394

8x2 + 16x + 10 – 394 = 0

8x2 + 16x – 384 = 0

8x2 + 2x – 48 = 0

x2 + 2x – 48 = 0

x2 + 8x – 6x – 48 = 0

x (x + 8) – 6 (x + 8) = 0

(x + 8) (x – 6) = 0

x + 8 = 0 or x – 6 = 0

x = -8 or x = 6

Then, first odd integer is (2x + 1) = (12 + 1) = 13

And next consecutive odd integer is (2x + 3) = (12 + 3) = 15 respectively

Thus, the required two consecutive odd integers are 13 & 15 respectively.

(3) The sum of the two numbers is 9 and the sum of their squares is 41. Taking one member as x, from all equation in x and solve it to find the numbers.

Solution:

Given that, the sum of two numbers is 9.

And sum of their squares is 41.

Let us consider the first number be ‘x’

Then next or second number be (9 – x).

Then, x2 + (9-x)2 = 41

x2 + 81 – 18x + x2 – 41 = 0

2x2 – 18x + 40 = 0

x2 – 9x + 20 = 0

x2 – 4x – 5x + 20 = 0

x (x – 4) – 5 (x – 4) = 0

(x – 4) (x – 5) = 0

x = 4 or x = 5

Then, the first number be x = 4

And second number be = 9 – x = 9 – 4 = 5.

Thus, the two required numbers are found to be 4 & 5.

(4) Five times a certain whole number is equal to three less than twice square of the number. Find the no.

Solution:

Let us consider ‘x’ be the required no.

From given condition, 5x = 2x2 – 3

2x2 – 3 – 5x = 0

2x2 – 5x – 3 = 0

2x2 – 6x + x – 3 = 0

2x (x – 3) + 1 (x – 3) = 0

(2x + 1) (x – 3) = 0

2x + 1 = 0 or x – 3 = 0

x = -1/2 or x = 3

Then, the required no. is found to be x = 3

(5) Sum of two natural numbers is 8 and the difference of their reciprocal is 2/15. Find the numbers.

Solution:

Let us consider the two natural no. x & y

From given condition,

x + y = 8 => y = 8 – x

1/x – 1/y = 2/15 —– (1)

Put y = 8 – x in equn(1) 1/x – 1/(8-x) = 2/15

(8-x)-x/x(8-x) = 2/15

(8-2x)/(8x-x2) = 2/15

120 – 30x = 16x – 2x2

2x2 – 16x – 30x + 120 = 0

2x2 – 46x + 120 = 0

x2 – 23x + 60 = 0

x2 – 20x – 3x + 60 = 0

x (x – 20) – 3 (x – 20) = 0

(x – 20) (x – 3) = 0

x – 20 = 0 or x – 3 = 0

x = 20 or x = 3

x ≠20 put x = 3 in x + y = 8

3 + y = 8

y = 5

Thus, the required two numbers are 3 & 5 respectively.

(7) There are three consecutive positive such that the sum of the square of the first & the product of the other two is 154. What are the integers?

Solution:

Let us consider the first integer be ‘x’.

Then second consecutive integer be ‘(x+1)’

Then third consecutive integer be ‘(x+2)’

From given condition,

x2 + (x+1) (x+2) = 154

x2 + (x2 + 2x + x + 2) = 154

2x2 + 3x + 2 = 154

2x2 + 3x + 2 – 154 = 0

2x2 + 3x – 152 = 0

2x2 + 19x – 16x – 152 = 0

x (2x + 19) – 8 (2x + 19) = 0

(2x + 19) (x – 8) = 0

2x + 19 = 0 or x – 8 = 0

x = -19/2 or x = 8

x = 8 is the first positive integer.

Then next two consecutive positive integers are 9 & 10 respect.

(9) In a certain positive fraction, the denominator is greater than the numerator by 3. If 1 is subtracted from both the numerator & denominator, the fraction is decreased by 1/14. Find the fraction.

Solution:

Let us consider the numerator of a fraction be ‘x’.

Then from given condition, denominator be (x + 3).

From given condition,

Fraction => x/x+3, x-1/x+12 = f – 1/14 42 = x2 + 5x + 6

x2 + 5x + 6 – 42 = 0

x2 + 5x – 36 = 0

x2 + 9x – 4x – 36 = 0

x (x + 9) – 4 (x + 9) = 0

(x + 9) (x – 4) = 0

x + 9 = 0 or x – 4 = 0

x = -9 or x = 4

Then denominator = x + 3

= 7

Thus, the required fraction is 4/7

Since, 4-1/7-1 = 3/6 & 4/7 – 3/6

= 24-21/42

= 3/42 = 1/14

(11) A two digit no. contains the bigger at ten’s place. Then product of the digits is 27 and the difference between two digits is 6. Find the number.

Solution:

Let us consider unit digit of that number is ‘x’.

Then ten’s digit of that number is x+6

Number = x + 10 (x + 6)

= x + 10x + 6

N = 11x + 60

From given condition,

x (x + 6) = 27

x2 + 6x – 27 = 0

x2 + 9x – 3x – 27 = 0

x (x + 9) – 3 (x + 9) = 0

(x + 9) (x – 3) = 0

x + 9 = 0 or x – 3 = 0

x = -9 or x = 3

x = -9 is not possible.

Thus put x = 3 in N = 11x + 60

= 11 (3) + 60 = 33 + 60 = 93

Hence, the required number is found to be 93.

(12) A two digit positive is such that the product of its digit is 5. If 9 is added to the number, the digits interchange their places. Find the number.

Solution:

Let us consider the two digit no. is xy = 10x + y

After reversing the digits = yx = 10y + x

From given condition,

10x + y + 9 = 10y + x and xy = 6

Put y = 6/x => 10x + 6/x + 9 = 10 (6/x) + x

10x2 + 6 + 9x = 60 + x2

9x2 + 9x – 54 = 0

x2 + x – 6 = 0

x2 + 3x – 2x – 6 = 0

x2 + 3x – 2x – 6 = 0

x (x + 3) – 2 (x + 3) = 0

(x + 3) (x – 2) = 0

x + 3 = 0 or x – 2 = 0

x = -3 or x = 2

The required value of x is 2.

Then y = 6/2 = 3

Hence, the required number is = 10x + y

= 20 + 3

N = 23

(14) A rectangular garden 10M by 16m is to be surrounded by a concrete walk of uniform width. Given that the area of the walk is 120 square meters, assuming the width of the walk to b x, form an equation in x and solve it to find the value of x.

Solution:

Given that, length of garden = 16m

Width of garden = 10m

Let us consider the width of the walk be ‘x’ m

Then, outer length = 16 + 2x and outer width = 10 + 2x

From given condition,

(16x + 2x) (10 + 2x) – 16 (10) = 120

160 + 32x + 20x + 4x2 – 160 – 120 = 0

4x2 + 52x – 120 = 0

x2 + 13x – 50 = 0

x2 + 15x – 2x – 30 = 0

x (x + 15) – 2 (x + 15) = 0

(x + 15) (x – 2) = 0

x + 15 = 0 or x – 2 = 0

x = -15 or x = 2

x = 2 is the required value.

(16) The perimeter of a rectangular plot is 180m and its area is 1800m2. Take the length of the plot as x m. Use the perimeter 180m to write the value of breadth in terms of x. Use the values of length, breadth and the area to write an equation in x. Solve the equation to calculate the length & breadth of the plot.

Solution:

Given that, the perimeter of rectangular filed = 180m.

Area = 1800m2

Let us consider length of rectangular plot be ‘x’

Then, perimeter = 2 (length + breadth)

180 = 2 (x + breadth)

Then, Area = length × breadth

1800 = x (90 – x)

90x – x2 = 1800

x2 – 90x + 1800 = 0

x2 – 60x – 30x + 1800 = 0

x (x – 60) – 30 (x – 60) = 0

(x – 60) (x – 30) = 0

x = 30 or x = 60

Hence, length = 60m

Breadth = 90 – 60 = 30m

(18) If the perimeter of a rectangular plot is 68m and the length of its diagonal is 26m, find its area.

Solution:

Given that, the perimeter of rectangular plot = 68m

Length of diagonal of plot = 26m

perimeter = 2 (l+b)

Let ‘x’ be the length of rectangular plot.

Then breadth = 34 – x

Now, Area = length × breadth

Area = x (34 – x) —- (1)

But, length2 + breadth2 = diagonal2

∵ By Pythagoras theorem

x2 + (34 – x)2 = 262

x2 + 1156 + x2 – 68x = 676

2x2 – 68x + 1156 – 676 = 0

2x2 – 68x + 480 = 0

x2 – 34x + 240 = 0

x2 – 24x – 10x + 240 = 0

x (x – 24) – 10 (x – 24) = 0

(x – 24) (x – 10) = 0

x – 24 = 0 or x – 10 = 0

x = 24 or x = 10

Thus, length = 24m

Breadth = 34 – 24 = 10m

Put in (1) => Area 24 × 10 = 240m2

Thus, the area of required rectangular plot is found to be 240m2.

(19) If the sum of two similar sides of a right – angle triangle is 17cm and the perimeter is 30cm, then find the area of the triangle.

Solution:

Given that, perimeter of the triangle = 30 cm

Let ‘x’ be the length of one of two small sides of triangle.

Then, other side = 17 – x

Area of a triangle = 1/2 (5×12) = 60/2 = 30cm2

Thus, the required area of a given triangle is 30cm2.

(21) Mohini wishes to fit three rods together in the shape of a tight triangle. If the hypotenuse is 2cm longer than the base and 4cm longer than the shortest side, find the lengths of the rod.

Solution:

Let us consider the length of hypotenuse be ‘x’ cm

Then base (x – 2) cm and shortest side = (x – 4) cm

x2 = (x – 2)2 + (x – 4)2

x2 = x2 – 4x + 4 + x2 – 8x + 16

x2 = 2x2 – 12x + 20

2x2 = 12x + 20 – x2 = 0

x2 – 10x – 2x + 20 = 0

x (x – 10) – 2 (x – 10) = 0

(x – 10) (x – 2) = 0

x – 10 = 0 or x – 2 = 0

x = 10 or x = 2

Here, hypotenuse = 10m

Base = 10 – 2 = 8

And shortest side = 10 – 4 = 6 cm

(22) In a P.T. display, 480 students are arranged in rows and columns. If there are 4 more students in each row than the number of rows, find the number of students in each row.

Solution:

Given that, the total no. of students = 480

Let ‘x’ be the number of students in each row.

Then, the no. of rows = 480/x

From given condition,

x = (480/x) + 4

x2 = 480 + 4x

x2 – 4x – 480 = 0

x2 – 24x + 20x – 480 = 0

x (x – 24) + 20 (x – 24) = 0

(x – 24) (x + 20) = 0

x = 24 or x = -20

Thus, the total no. of students in each row = 24

(24) At an annual function of a school, each student gives the gift to every other student. If the number of gifts is 1980, find the number of students.

Solution:

Let us consider the total no. of students ax ‘x’

Then no. of gifts given = x – 1

Total no. of gifts = x (x – 1)

From given condition,

x (x – 1) = 1980

x2 – x – 1980 = 0

x2 – 45x + 44x – 1980 = 0

x (x – 45) + 44 (x – 45) = 0

(x – 45) (x + 44) = 0

x – 45 = 0, x + 44 = 0

x = 45 or x = -44

Thus, the total no. of students = 45

(26) The speed of an express train is x km/hr and the speed of an ordinary train is 12 km/hr less than that of the express train. If the ordinary train takes one hour longer than the express train to cover a distance of 240km, find the speed of the express train.

Solution:

Let us consider the speed of the express train = x km

Then, the speed of ordinary train = (x – 12) km

Time taken to cover 240 km by the express.

Then, time require to cover for each train is 240/x & 240/(x-12) respectively.

From given condition,

240x – 12 – 240x = 1

240x – 240 (x – 12) (x – 12) x = 1

240x – 240 (x – 12) = x (x – 12)

x2 – 12x – 2880 = 0

(x – 60) (x + 48) = 0

x = 60 km/hr

Thus, the speed of the express train is 60 km/hr.

(30) An aeroplane flying with a wind of 30 km/hr takes 40 minutes less to fly 3600 km, than what it would have taken to fly against the same wind. Find the planes speed of flying in still air.

Solution:

Let us consider the speed of the plane in still air = x km/hr

And speed of wind = 30km/hr

Distance = 3600/km

Hence, Time taken with the wind = 3600/x+30

Time taken against the wind = 3600/x-30

From given condition,

3600/(x-30) – 3600/(x+30) = 40 minutes = 2/3 hour

3600 (1/x-30 – 1/x+30) = 2/3 2x2 – 1800 = 3×3600×60

2x2 – 1800 = 64,8000

2x2 – 1800 – 648000 = 0

2x2 – 649800 = 0

x2 – 324900 = 0

x2 – (570)2 = 0

(x + 570) (x – 570) = 0

x + 570 = 0 or x – 570 = 0

x = -570 or x = 570

Thus, the speed of the plane in still air = 570 km/hr

(32) A boat can cover 10km up the stream and 5km down the stream in 6 hours. If the speed of the stream is km/hr. Find the speed of the boat in still water.

Solution:

Given that, distance up stream = 10 km

Distance down stream = 5 km

Total time taken = 6 hours

Speed of the stream = 1.5 km/hr

Let us consider the speed of the boat in still water is x km/hr from given condition,

x/x-1.5 + 5/x+1.5 = 6

10x + 15 + 5x + 5x – 7.5 = 6 (x – 15) (x + 15)

15x + 7.5 = 6 (x2 – 2.25)

15x + 7.5 = 6x2 – 13.5

6x2 – 15x – 13.5 – 7.5 = 0

6x2 – 15x – 21 = 0

2x2 – 5x – 7 = 0

2x2 – 7x + 2x – 7 = 0

x (2x – 7) + 1 (2x – 7) = 0

(2x – 7) (x + 1) = 0

2x = 7 or x = -1

x = 7/2

Thus, x = 7/2 = 3.5

Thus, the speed of the boat in still water is found to be 3.5 km/hr.

(36) A trader buys x articles for a total cost of Rs 600.

(i) Write down the cost of one article in terms of x.

If the cost per article were Rs 5 more, the number of articles that can be bought for Rs 600 would be four less.

(ii) Write down the equation in x for the above situation and solve it to find x.

Solution:

Given that, the total cost of x article = 600 Rs.

No. of articles bought = x

Then, Cost of one article = 600/x

From given condition,

600/(x-4) – 600/x = 5

=> x2 – 4x – 480 = 0

x2 – 24x + 20x – 480 = 0

x (x – 24) + 20 (x – 24) = 0

(x – 24) (x + 20) = 0

x = 24 or x = -20

Thus, the total no. of articles are found to be 24.

(39) The hotel bill for a number of people for an overnight story is Rs 4800. If there were 4 more, the bill each person had to pay would have reduced by Rs. 200. Find the no. of people staying overnight.

Solution:

Let us consider the no. of people staying overnight be ‘x’.

Amount of bill = 4800 Rs.

=> Bill for each person = 4800/x

From second condition,

The total no. of people = x + 4

Then bill for each person = 4800/(x+4)

From the given condition, we can write

4800/x – 4800/(x+4) = 200

4800 [1/x – 1/(x+4)] = 200

=> 19200 = 200x2 + 800x

200x2 + 800x – 19200 = x

x2 + 4x – 96 = 0

x2 + 12x – 8x – 96 = 0

x (x + 12) – 8 (x + 12) = 0

(x + 12) (x – 8) = 0

x = -12 or x = 8

Thus, the total no. people staying in hotel ax found to be 8.

(42) The sum of the ages of Vivek and his younger brother Amit is 47 years. The product of their ages in years is 550. Find their ages.

Solution:

Let us consider the vivek’s present age be ‘x’ years.

Then, His brother’s age will be (47 – x) years.

From given condition,

x (47 – x) = 550

47x – x2 = 550

x2 – 47x + 550 = 0

x2 – 25x – 22x + 550 = 0

x (x – 25) – 22 (x – 25) = 0

(x – 22) (x – 25) = 0

x = 22 or x = 25

When x = 25 then 47 – x = 47 – 25 = 22

When x = 22 then 47 – x = 47 – 22 = 25

Hence, the Vivek’s age is found to be x = 25 years. His younger brother’s age be 22 years.

(44) Two years ago, a man’s age was three times the square of his daughter’s age. Three years hence, his age will be four times his daughter’s age. Find their present ages.

Solution:

Given that, 2 years ago

Let us consider the age of daughter be ‘x’.

Then age of the man = 3x2

At present, the age of daughter be (x + 2) and age of man = (3x2 + 2)

After 3 years again, the age of daughter = x + 2 + 3

= (x + 5)

And the age of man = 3x2 + 2 + 3 = (3x2 + 5)

From given condition,

3x2 + 5 = 4 (x + 5)

3x2 + 5 = 4x + 20

3x2 – 4x + 5 – 20 = 0

3x2 – 4x – 15 = 0

3x2 – 9x + 5x – 15 = 0

3x (x – 3) + 5 (x – 3) = 0

(x – 3) (3x + 5) = 0

x = 3 or x = -5/3

If x = 3 then present age of man = 3x2 + 2

Present age of man = 3x2 + 2 = 27 + 2 = 29 years

And age of daughter = x + 2 = 3 + 2 = 5 years.

(45) The length of the hypotenuse of a right-angled triangle exceeds the length of one side by 2 cm and exceeds twice the length of another side by 1cm. Find the length of each side. Also, find the perimeter and the area of the triangle.

Solution:

Let us consider the length of one side of triangle be ‘x’ cm.

And of another side be y cm.

Then hypotenuse = x + 2 and (2y + 1)

∴ x + 2 = 2y + 1

x – 2y – 1 + 2 = 0

x – 2y = -1

x = 2y – 1 —- (1)

By Pythagoras theorem,

x2 + y2 = (2y + 1)2

x2 + y2 = 4y2 + 4y + 1

(2y – 1)2 + y2 = 4y2 + 4y + 1

4y2 – 4y + 1 + y2 – 4y2 – 4y – 1 = 0

y2 – 8y = 0

y (y – 8) = 0

y = 8 put in (1) = x = 2y – 1

= 16 – 1

x = 15

Hence, length of one side = 15 cm

Length of another side = 8 cm

Hypotenuse = x + 2 = 15 + 2 = 17 cm

Area of triangle = 1/2 × (8) (15) = 60 cm2

Updated: February 6, 2023 — 2:01 pm