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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 :

JEE · Math · previous-year question

  1. A.$$x^2+8 x+12=0$$correct
  2. B.$$2 x^2+11 x+12=0$$
  3. C.$$4 x^2+14 x+12=0$$
  4. D.$$x^2+10 x+16=0$$

Answer

A. $$x^2+8 x+12=0$$

Explanation

Given that the roots of the quadratic equation are $2$ and $6$, we can use Vieta's formulas which relate the coefficients of the polynomial to sums and products of its roots. The given quadratic equation is: $$a x^2 + b x + 1 = 0$$ By Vieta's formulas, the sum of the roots is: $$2 + 6 = -\frac{b}{a}$$ So: $$8 = -\frac{b}{a} \Rightarrow b = -8a$$ And the product of the roots is: $$2 \times 6 = \frac{1}{a} \Rightarrow 12 = \frac{1}{a} \Rightarrow a = \frac{1}{12}$$ Therefore, $b = -8a = -8 \left( \frac{1}{12} \right) = -\frac{2}{3}$. Given the roots of the new quadratic equation are: $$\frac{1}{2a+b}$$ and $$\frac{1}{6a+b}$$ We know $a = \frac{1}{12}$ and $b = -\frac{2}{3}$, so: $$2a + b = 2 \left(\frac{1}{12}\right) - \frac{2}{3} = \frac{1}{6} - \frac{2}{3} = \frac{1 - 4}{6} = -\frac{3}{6} = -\frac{1}{2}$$ and: $$6a + b = 6 \left(\frac{1}{12}\right) - \frac{2}{3} = \frac{1}{2} - \frac{2}{3} = \frac{3 - 4}{6} = -\frac{1}{6}$$ Thus, the roots of the new quadratic equation are: $$\frac{1}{-\frac{1}{2}} = -2$$ and: $$\frac{1}{-\frac{1}{6}} = -6$$ The new quadratic equation with roots $-2$ and $-6$ can be formulated as: $$x^2 - (\text{sum of roots}) x + (\text{product of roots}) = 0$$ The sum of the roots is: $$-2 + (-6) = -8$$ The product of the roots is: $$(-2) \times (-6) = 12$$ Thus, the quadratic equation becomes: $$x^2 - (-8)x + 12 = x^2 + 8x + 12 = 0$$ Hence, the correct option is: Option A $$x^2 + 8 x + 12 = 0$$

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