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Question: Let \(\frac{\alpha}{\alpha –1}\) and \(\frac{\beta}{\beta –1}\) be roots of x<sup>2</sup> + ax + b =...

Let αα–1\frac{\alpha}{\alpha –1} and ββ–1\frac{\beta}{\beta –1} be roots of x2 + ax + b = 0 then 1α\frac{1}{\alpha} and

1β\frac{1}{\beta} are roots of :

A

bX2 + aX + 1 = 0

B

bX2 – aX + 1 = 0

C

bX2 + (a + 2b)X + a + b + 1 = 0

D

bX2 – (a + 2b)X + a + b + 1 = 0

Answer

bX2 – (a + 2b)X + a + b + 1 = 0

Explanation

Solution

αα–1\frac{\alpha}{\alpha –1} or ββ–1\frac{\beta}{\beta –1} = x

α–1α\frac{\alpha –1}{\alpha} or β–1β\frac{\beta –1}{\beta} = 1x\frac{1}{x}

1 – 1α\frac{1}{\alpha} or 1 – 1β\frac{1}{\beta} = 1x\frac{1}{x}

–1α\frac{1}{\alpha} or – 1β\frac{1}{\beta} = 1x\frac{1}{x}–1

1α\frac{1}{\alpha} or 1β\frac{1}{\beta} = 1 – 1x\frac{1}{x}

∴ X = 1– 1x\frac{1}{x} ⇒ 1x\frac{1}{x} = 1– X

x = 11–X\frac{1}{1–X}

Replace x with 11–X\frac{1}{1–X} in present equation

(11–X)2\left( \frac{1}{1–X} \right)^{2} + a(11–X)\left( \frac{1}{1–X} \right) + b = 0

b(1– X)2 + a(1 – X) + 1 = 0

bX2 – (2b + a) X + b + a + 1= 0