If 2 and 6 are the roots of the equation a x 2 + b x + 1 = 0 , then the quadratic equation, whose roots are 1 2 a + b and 1 6 a + b , is [2024]

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Solution

Q.
If 2 and 6 are the roots of the equation $a x^{2} + b x + 1 = 0$ , then the quadratic equation, whose roots are $\frac{1}{2 a + b}$ and $\frac{1}{6 a + b},$ is [2024]

1 $2 x^{2} + 11 x + 12 = 0$

2 $x^{2} + 8 x + 12 = 0$

3 $4 x^{2} + 14 x + 12 = 0$

4 $x^{2} + 10 x + 16 = 0$

Ans.
(2)

2 and 6 are the roots of the equation $a x^{2} + b x + 1 = 0$

$∴$ Sum of roots = $\frac{- b}{a} = 8 \Rightarrow b = - 8 a$

Product of roots = $\frac{1}{a} = 12 \Rightarrow a = \frac{1}{12}$

$∴$ $b = \frac{- 2}{3} .$ Now, $\frac{1}{2 a + b} = \frac{1}{2 \times \frac{1}{12} - \frac{2}{3}} = - 2$

and $\frac{1}{6 a + b} = \frac{1}{6 \times \frac{1}{12} - \frac{2}{3}} = - 61$

$∴$ Required equation is $(x + 2) (x + 6) = 0$

i.e., $x^{2} + 8 x + 12 = 0$

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