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The quadratic equation (1 + a2)x2 + 2abx + (b2 - c2) = 0 has only one root. What is the value of c2(1 + a2)?
- Aa2
- Bb2
- Cc2
- Dab
Answer not set
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CCommunity
(1 + a2)x2 + 2abx + (b2 - c2)= 0
Comparing on Ax2 + 2abx + C = 0
A = 1 + a2, B = 2ab & C = (b2 - c2)
Now, B2 - 4AC = 0
⇒ (2ab)2 - 4 × (1 + a2) × (b2 - c2) = 0
⇒ 4a2b2 - 4(b2 - c2 + a2b2 - a2c2) = 0
⇒ 4a2b2 - 4b2 + 4c2 - 4a2b2 + 4 a2c2 = 0
⇒ - b2+ c2 + a2c2 = 0
⇒ c2 + a2c2 = b2
∴ c2 (1 + a2) = b2
Comparing on Ax2 + 2abx + C = 0
A = 1 + a2, B = 2ab & C = (b2 - c2)
Now, B2 - 4AC = 0
⇒ (2ab)2 - 4 × (1 + a2) × (b2 - c2) = 0
⇒ 4a2b2 - 4(b2 - c2 + a2b2 - a2c2) = 0
⇒ 4a2b2 - 4b2 + 4c2 - 4a2b2 + 4 a2c2 = 0
⇒ - b2+ c2 + a2c2 = 0
⇒ c2 + a2c2 = b2
∴ c2 (1 + a2) = b2
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CCommunity
$b^2$