Telangana SCERT Solution Class X (10) Maths Chapter 5 Quadratic Equations Exercise 5.4
Quadratic Equations
Exercise – 5.4
(Q1) Find the nature of the roots of the following quadratic equation. If real roots exist, find them.
(i) 2x2 – 3x + 5 = 0
=> Solution: Given equation is 2x2 – 3x + 5 = 0
Comparing with ax2 + bx + c = 0
Here, a = 2, b = – 3, c = 5
b2 – 4ac = (-3)2 – 4(2) (5)
= 9 – 40
= – 31 <0
∴ There are no real roots.
(ii) 3x2 – 4√3x + 4 = 0
=> Solution: Given equation is 3x2 – 4√3x + 4 = 0
Comparing with ax2 + bx + c = 0
Here, a = 3, b = 4√3, c = 4
b2 – 4ac = (- 4√3)2 – 4 (3) (4)
= 48 – 48
= 0
The roots are real and equal.
Roots = -b/2a, -b/2a
= (-4√3)/2X3, – (-4√3)/2X3
= 4√3/2X3, 4√3/2X3
= 2√3/3, 2√3/3
Roots = 2√3/3, 2√3/3
(Q2) Find the values of K for each of the following quadratic equations, so that they have to equal roots
(1) 2x2 + kx + 3 = 0
=> Solution: Given equation is 2x2 + kx + 3 = 0 —– (1)
Comparing with ax2 + bx + c = 0
Here, a = 2, b = K, c = 3.
Since equation (1) has equal roots
b2 – 4ac = 0
K2 – 4(2) (3) = 0
K2 – 24 = 0
K2 = 24
K = ± √24
K = ± 2√6
(ii) Kx (x – 2) + 6 = 0 (k ≠ 0)
=> Solution: Given equation is Kx (x – 2) +6 = 0
Kx2 – 2Kx + 6 = 0 —– (1)
Comparing with ax2 + bx + c = 0
Here, a = k, b = 2k, c = 6
Since equation (1) has equal roots,
b2 – 4ac = 0
(-2K)2 – 4(K) (6) = 0
4K2 – 24K = 0
4K (K – 6) = 0
K ( K – 6) = 0
Since K ≠ 0, K – 6 = 0
∴ K = 6
x = ± √400
x = ± 20
Since ‘x’ is the breadth, it must be positive.
∴ x = 20
If x = 20 then 2x = 2 X 20 = 40
Yes, it is possible to design rectangular mango grove
It’s breadth = 20m and length = 40m.
(Q4) The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48. Is the above situation possible? If so, determine their present ages.
=> Solution: The sum of the ages of two friend = 20 years
Let, the age of first person be ‘c’ years.
Then the age of second person = (20 – x) years
Before 4 years, age of first person = (x – 4) years
And the age of second person = 20 – x – 4
= (16 – x) years
Since their product = 48
i.e. (x – 4) (16 – x) = 48
x (16 – x) – 4(16 – x) = 48
16x – x2 – 64 – 4x = 48
12x – x2 – 64 – 48 = 0
12x – x2 – 112 = 0
x2 – 12x + 112 = 0
Comparing with ax2 + bx + c = 0
Here, a = 1, b = – 12, c = 112
b2 – 4ac = (-12)2 – 4(1) (112)
= 144 – 448
= – 304 <0
∴ Roots are not real
∴ The situation is not possible
(Q5) Is it possible to design a rectangular park of perimeter 80m and area 400 m2? If so, find its length and breadth comment on your answer.
i.e. 2 (l + b) = 80
l + b = 80/2 = 40
Let, the breadth of the park be ‘x’ m
Then its length, l = 40 – x (∵ from (1))
But its area = 400m
I.e. l X b = 400
(40 – x) x = 400
40x – x2 = 400
x2 – 40x + 400 = 0 ——– (2) P = 80m, A = 400m
Comparing with ax2 + bx + c = 0
Here, a = 1, b = – 40, c = 400
b2 – 4ac = (-40)2– 4(1) (400)
= 1600 – 1600
= 0
∴ Roots are real and equal.
∴ It is possible to design such rectangular part.
∴ x = -b/2a, -b/2a
= -(-40)/2X1, -(-40)/2X1
= 40/2, 40/2
X = 20, 20
∴ Since, x = 20, 40 – x = 40-20 = 20
∴ It’s breadth = 20m and length = 20m
Since breadth and length are equal, it is a square park.
Here is your solution of Telangana SCERT Class 10 Math Chapter 5 Quadratic Equations Exercise 5.4
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