Question

If $\alpha$ and $\beta$ are the roots of the equation $x^2+5x-7=0$. Then a equation with roots
$\frac{1}{\alpha }and\frac{1}{\beta }$
is .

• A. $7x^2-5x-1=0$
• B. $7x^2+5x+1=0$
• C. $7x^2-5x+1=0$
• D. $7x^2+5x-1=0$

Answer 1

The correct option is A $7x^2-5x-1=0$

The equation,$x^2+5x-7=0$ has roots, $\alpha$ and $\beta$$\Rightarrow \alpha +\beta =-5and\alpha \times \beta =-7......\left(1\right]$
when the roots of a equation are given then its equation is of the form-
$x^2$- (sum of its roots)x + (product of its roots)=0
$\therefore equationwithroots,\frac{1}{\alpha }and\frac{1}{\beta }is{x}^2-(\frac{1}{\alpha }+\frac{1}{\beta })x+\frac{1}{\alpha }\times \frac{1}{\beta }=0x^2-\left(\frac{\alpha +\beta }{\alpha \times \beta }\right)x+\frac{1}{\alpha \times \beta }=0$
using [1] we get,
$7x^2-5x-1=0$

Alternatively Solution:
If the roots of the new equation are reciprocal of the original equation, then we will replace x with $\frac{1}{x}$ in the original equation.
Hence, a new equation formed will be:
$(\frac{1}{x}{)}^2+5(\frac{1}{x})-7=0\Rightarrow \frac{1+5x-7x^2}{x^2}=0\Rightarrow 1+5x-7x^2=0\Rightarrow 7x^2-5x-1=0$