If 2 and 6 are the roots of the equation a x^2+b x+1=0, then the quadratic equation, whose…

Question

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 :

Accepted Answer

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

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

Solutions · 1