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Question: If α, β are roots of equation a*x*<sup>2</sup> + b*x* + c = 0 then the equation a(*x* – 2)<sup>2</s...

If α, β are roots of equation ax2 + bx + c = 0 then the equation

a(x – 2)2 – b(x – 1) (x – 2) + c(x – 1)2 = 0 has roots-

A

α−2α−1\frac{\alpha - 2}{\alpha - 1}, β−2β−1\frac{\beta - 2}{\beta - 1}

B

α+2α+1\frac{\alpha + 2}{\alpha + 1}, β+2β+1\frac{\beta + 2}{\beta + 1}

C

α−2α+1\frac{\alpha - 2}{\alpha + 1}, β−2β+1\frac{\beta - 2}{\beta + 1}

D

α+2α−1\frac{\alpha + 2}{\alpha - 1}, β+2β−1\frac{\beta + 2}{\beta - 1}

Answer

α+2α+1\frac{\alpha + 2}{\alpha + 1}, β+2β+1\frac{\beta + 2}{\beta + 1}

Explanation

Solution

a (−(x−2)(x−1))2\left( \frac{- (x - 2)}{(x - 1)} \right)^{2}+ b(−(x−1)(x−2))\left( \frac{- (x - 1)}{(x - 2)} \right) + c = 0

Now replace −(x−2)(x−1)\frac{- (x - 2)}{(x - 1)} by α 

Then α = −(x−2)x−1\frac{- (x - 2)}{x - 1}

⇒ αx – α = –x + 2

⇒ x(α + 1) = α + 2

⇒ x = (α+2α+1)\left( \frac{\alpha + 2}{\alpha + 1} \right)

Roots (α+2α+1,β+2β+1)\left( \frac{\alpha + 2}{\alpha + 1},\frac{\beta + 2}{\beta + 1} \right).