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For what values of a, the quadratic equation 9x^2 – 3ax + 1 = 0 has real and equal roots?

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Question

For what values of a, the quadratic equation 9x2 – 3ax + 1 = 0 has real and equal roots?

Sum
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Solution

Given: 9x2 – 3ax + 1 = 0

Step-wise calculation:

1. For real and equal roots the discriminant must be zero:

D = b2 – 4ac = 0

2. Here A = 9, B = –3a, C = 1

So, D = (–3a)2 – 4 × 9 × 1 

= 9a2 – 36

= 9(a2 – 4)

3. Set D = 0

⇒ 9(a2 – 4) = 0 

⇒ a2 – 4 = 0

⇒ a2 = 4 

⇒ a = ±2

The quadratic has real and equal roots for a = 2 or a = –2.

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Chapter 4: Quadratic Equations - EXERCISE 4C [Page 203]

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R.S. Aggarwal Mathematics [English] Class 10
Chapter 4 Quadratic Equations
EXERCISE 4C | Q 31. | Page 203
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