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Find the values of k for which the following equation has equal roots: x^2 – 2(5 + 2k)x + 3(7 + 10k) = 0

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Question

Find the values of k for which the following equation has equal roots:

x2 – 2(5 + 2k)x + 3(7 + 10k) = 0

Sum
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Solution

Given: x2 – 2(5 + 2k)x + 3(7 + 10k) = 0

Step-wise calculation:

1. Compare with ax2 + bx + c = 0: 

a = 1, b = –2(5 + 2k) = –10 – 4k, c = 3(7 + 10k) = 21 + 30k

2. For equal (repeated) roots the discriminant D = b2 – 4ac must be 0 (D = 0).

3. Compute D: b2 = (–10 – 4k)2 

= 100 + 80k + 16k2

4ac = 4(21 + 30k)

= 84 + 120k

D = b2 – 4ac

= (100 + 80k + 16k2) – (84 + 120k)

= 16k2 – 40k + 16

4. Set D = 0 and simplify:

16k2 – 40k + 16 = 0

⇒ Divide by 4

⇒ 4k2 – 10k + 4 = 0

5. Solve 4k2 – 10k + 4 = 0 using quadratic formula:

`k = (10 ± sqrt(100 - 64))/8` 

= `(10 ± 6)/8` 

So, `k = 16/8 = 2` or `k = 4/8 = 1/2`.

The equation has equal roots for k = 2 or k = `1/2`.

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Chapter 5: Quadratic Equation - EXERCISE 5C [Page 61]

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R.S. Aggarwal Mathematics [English] Class 10 ICSE
Chapter 5 Quadratic Equation
EXERCISE 5C | Q 16. | Page 61
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