Selina Concise Class 10 Math Chapter 5 Quadratic Equations Exercise 5C Solutions
Quadratic Equations Exercise 5C
(Q1) Solve equations, number 1 to 20, given below, using factorization method
(i) x² – 10x – 24 = 0
= Solution:
The given equation is x² – 10x – 24 = 0
Find factor of (-24)
= x² – 10 – 24 = 0
= x² – 6x – 4x – 24 = 0
= x (x – 6) – 4 (x – 6) = 0
= (x – 6) (x – 4) = 0
= x – 6 = 0 or x – 4 = 0
(-6 – 4 = -10)
= x = 6 or x = 4
∴ The factors of given equation are x = (6, 4)
(2) x² – 16 = 0
= Solution:
The given equation is x² – 16
Find factor of (-16)
= x² – 16 = 0
= x² – 4x – 4x – 16 = 0
= x (x – 4) – 4 (x – 4) = 0
= (x – 4) (x – 4) = 0
= x- 4 = 0 or (-4 -4 = -8)
x – 4 = 0
= x = 4 or x = 4
∴ The solution is x = (4, 4)
(3) 2x² – ½ x = 0
= Solution:
The given equation is 2x² – ½ x = 0
Multiply by 2,
2 × 2x² – 2 × ½ x = 2 × 0
4x² – x = 0
x (4x – 1) = 0
x = 0 or (4x – 1) = 0
4x – 1 = 0
4x = 1
x = ¼
The solution is x = (0, ¼)
(4) x (x – 5) = 24
= Solution:
The given equation is x (x – 5) = 24
= x² – 5x = 24
= x (x – 5) = 24
= x (x – 5) – 24 = 0
= x² – 5x – 24 = 0
Find factor of ‘-24’
= x² – 5x – 24 = 0
= x² – 8x + 3x – 24 = 0
= x (x – 8) + 3 (x – 8) = 0
= (x – 8) (x + 3) = 0
= x – 8 = 0 or x + 3 = 0
(-8+3 = -5)
= x = 8 or x = -3
∴ The solution is x = (8, -3)
(5) 9/2 x = 5+x²
= Solution:
The given equation is
9/2 x = 5 + x²
Multiply by ‘2’ on both sides
2 × 9/2 x = 2 × (5 + x²)
9x = 10 + 2x²
2x² – 9x + 10 = 0
Find factor of (10 × 2 = 20)
2x² – 9x + 10 = 0
2x² – 5x – 4x + 10 = 0
x (2x – 5) – 2 (2x – 5) = 0
(-5-4 = -9)
(2x – 5) (x – 2) = 0
2x – 5 = 0 or x – 2 = 0
2x = 5 or x = 2
x = 5/2
(6) 6/x = 1+x
= Solution:
The given equation is 6/x = 1+x
Multiply both sides by ‘x’
6/x × x = x (1+x)
6 = x + x²
x² + x – 6 = 0
Find factor (-6)
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 (+3 -2= 1)
x = -3 or x = 2
(7) x = (3x + 1/4x)
= Solution:
The given equation is x = (3x+1/4x)
Multiply by ‘4x’ on both side,
(4x) × x = 4x × (3x+1/4x)
4x² = 3x+1
4x² – 3x – 1 = 0
Find factor (4× (-1) = -4)
4x² – 3x – 1 = 0
4x² – 4x + x – 1 = 0
4x (x – 1) +1 (x – 1) = 0
(x – 1) (4x + 1) = 0
(x – 1) or 4x + 1 = 0
(-4+1 = -3)
x = 1 or 4x = -1
x = -1/4
(8) x + 1/x = 2.5
= Solution:
The given equation is
x + 1/x = 2.5
x + 1/x = 5/2 (2.5 = 5/2)
Multiply by ‘x’ on both sides,
x × x + x × 1/x = x × 5/2
x² + 1 = x × 5/2
Multiply by ‘2’ on both sides,
2x² + 2 = x × 2 × 5/2
2x² + 2 = 5x
2x² – 5x + 2 = 0
Find factor (2 × 2 = 4)
2x² – 5x + 2 = 0
2x² – 4x – 1x + 2 = 0
2x (x – 2) -1 (x – 2) = 0
(x – 2) (2x – 1) = 0
x – 2 =0 or 2x – 1 = 0 (-4-1 = -5)
x = 2 or 2x = 21
x = ½
(9) (2x – 3)²
= Solution:
The given equation is
(2x – 3)² = 49
(2x)² – 2× (2x) (-3) + (3)² – 49 [∵ (a-b)² = a² – 2ab + b²]
4x² + 12x + 9 = 49
4x² + 12x + 9 – 49 = 0
4x² + 12x – 40 = 0
Divide ‘4’
4x²/4 + 12x/4 – 40/4 = 0
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
(+5-2 =3)
x = -5 or x = 2
(10) 2(x² – 6) = 3 (x – 4)
= Solution:
The given equation is 2 (x² – 6) = 3 (x – 4)
= 2x² – 12 = 3x – 12
= 2x² – 3x – 12 + 12 = 0
2x² – 3x – 0 = 0
2x² – 3x = 0
= x (2x – 3) = 0
= x = 0 or 2x – 3 = 0
2x = 3
x = 3/2
(11) (x + 1) (2x + 8) = (x + 7) (x + 3)
= Solution:
The given equation is (x + 1) (2x + 8) = (x + 7) (x + 3)
= x × (2x + 8) + 1 (2x + 8) = x (x + 3) + 7 (x + 3)
= 2x² + 8x + 2x + 8 = x² + 3x + 7x + 21
= 2x² – x² + 8x – 3x + 2x – 7x + 8 – 21 = 0
= x² + 5x – 5x – 13 = 0
= x² + 0 – 13 = 0
= x² – 13 = 0
= x² = 13 (Take square root on both side)
= x = ± √13
= x = + √13 or x = – √13
(12) x² – (a + b) x + ab = 0
= Solution:
The given equation is x² – (a + b) x + ab = 0
x² – ax + bx + ab = 0
x (x – a) – b (x – a) = 0
(x – a) (x – b) = 0
x – a = 0 or x – b = 0
x = a or x = b
(13) (x+3)² – 4 (x + 3) – 5 = 0
= Solution:
The given equation is (x + 3)² – 4 (x + 3) – 5 = 0
x² + 2× x ×3 + (3)² – [4x + 12] – 5 = 0
x² + 6x + 9 – 4x – 12 – 5 = 0
x² + 2x – 8 = 0
x² + 4x – 2x – 8 = 0
x (x + 4) -2 (x + 4) = 0
(x + 4) (x – 2) = 0
x + 4 = 0 or x – 2 = 0 (4-2 = 2)
x = -4 or x = 2
(14) 4 (2x – 3)² – (2x – 3) – 14 – 0
= Solution:
The given equation is 4 (2x – 3)² – (2x – 3) -14 = 0
Substitute 2x – 3 = P
4P² – P – 14 = 0
4P² – P – 14 = 0
4P² – 8P + 7P – 14 = 0 (4× (-14) = -56)
4P (P – 2) + 7 (P – 2) = 0
(P – 2) (4P + 7) = 0
P – 2 = 0 or (4P + 7) = 0
P = 2 or 4P + 7 = 0
(-8+7 = -1)
4P = -7
P = -7/4
But P = 2x – 3 and P = 2x – 3
2 = 2x – 3
2+3 = 2x
5 = 2x
5/2 = x
∴ x = 5/2
-7/4 = 2x – 3
-7 = 4 (2x – 3)
-7 = 8x – 12
-7+12 = 8x
5 = 8x
5/8 = x
(15) 3x-2/2x-3 = 3x-8/x+4
= Solution:
The given equation is 3x-2/2x-3 = 3x-8/x+4
Cross multiply on both side,
(3x – 2) (x + 4) = (3x – 8) (2x – 3)
3x (x + 4) – 2 (x + 4) = 3x (2x – 3) – 8 (2x – 3)
3x² + 12x – 2x – 8 = 6x² – 9x – 16x + 24
3x²- 6x² + 10x + 9x + 16x – 8 – 24 = 0
-3x² + 35x – 32 = 0
3x² – 35x + 32 =0 (multiply by (-1))
3x² – 32x – 3x + 32 = 0
x (3x – 32) -1 (3x – 32) = 0
(3x – 32) (x – 1) = 0
3x – 32 = 0 or x – 1 = 0
3x = 32 or x = 1
x = 32/3
(16) 2x² – 9x + 10 = 0, when
(i) x ∈ N (ii) x ∈ Q
= (i) Solution:
The given equation is 2x² – 9x + 10 = 0
= 2x² – 9x + 10 = 0
= 2x² – 5x – 4x + 10 = 0
= x (2x – 5) – 2 (2x – 5) = 0
(2x – 5) (x – 2) = 0
= 2x – 5 = 0 or x – 2 = 0 (-5-4 = -9)
= 2x = 5 or x = 2
= x = 5/2
(i) when x∈N
(N – natural number)
So, x = 2 is the solution
x = 5/2 is not the solution because its p/q form means rational number.
(ii) When x ∈ Q
(Q – Rational numbers)
So, x = 5/2, 2 are the solution)
(17) x-3/x+3 + x+3/x-3 = 2 ½
= Solution:
The given equation is
x-3 / x+3 + x+3 / x-3 = 2 ½
= take L.C.M
= (x-3)² + (x + 3)²/(x + 3) (x – 3) = 4+1/2
= x²- 2× x ×(+3) + (-3)² + x² + 2× x ×3 + (3)²/x (x-3) + 3 (x – 3) = 5/2
= 2x²- 6x + 9 + 6x + 9/x² – 3x + 3x – 9 = 5/2
= 2x² + 18/x²-9 = 5/2
= 2 (2x² + 18) = 5 (x² – 9)
= 4x² + 36 = 5x² – 45
= 4x² – 5x² + 36 + 45 = 0
= -x² + 81 = 0
= x² – 81 = 0 (multiply by (-1))
= x² – (9)² = 0
= (x – 9) (x + 9) = 0 (a² – b² = (a – b) (a + b)
= x – 9 = 0 or x + 9 = 0
= x = 9 or x = -9