Question

Obtain a quadratic equation whose roots are reciprocals of the roots of the equation $x^2-3x-4=0$.

• A. $4x^2-3x+1=0$
• B. $4x^2+3x+1=0$
• C. $4x^2-3x-1=0$
• D. $4x^2+3x-1=0$

Answer 1

The correct option is D $4x^2+3x-1=0$

The given equation is $x^2-3x-4=0$

$x^2-3x-4=0$

$\Rightarrow$ $x^2-4x+x-4=0$

$\Rightarrow$ $x(x-4)+1(x-4)=0$

$\Rightarrow$ $(x-4)(x+1)=0$

$\Rightarrow$ $x=4$ and $x=-1$

Taking reciprocal of roots we get,

$\Rightarrow$ $x=\frac{1}{4}$ and $x=-1$

The required equation,

$\Rightarrow$ $(x-\frac{1}{4})(x+1)=0$

$\Rightarrow$ $x^2+x-\frac{x}{4}-\frac{1}{4}=0$

$\Rightarrow$ $4x^2+4x-x-1=0$ [ Multiplying both sides by

$4$ ]

$\therefore$ $4x^2+3x-1=0$