Question

Two students Ragini and Gourav were asked to solve a quadratic equation $a{x}^2+bx+c=0,a\ne 0$ Ragini made some mistake in writing b and found the roots as 3 and $-\frac{1}{2}$ Gourav too made mistake in writing c and found the roots -1 and $-\frac{1}{4}$ The correct roots of the given equation should be

• A. $-2,$ $\frac{3}{4}$
• B. $3,-1$
• C. $-\frac{1}{2}$, -1
• D. $3,$ $-\frac{1}{4}$

Answer 1

The correct option is A $-2,$ $\frac{3}{4}$

Given: Ragini found roots as $3,-\frac{1}{2}$ when copied the wrong coefficient of $x$ and Gourav found roots as $-1,-\frac{1}{4}$ when copied wrong constant term.

To find the correct roots of the given equation

Sol: Viete's formula for the roots $x_1$ and $x_2$ of equation $a{x}^2+bx+c=0$: $x_1+x_2=-\frac{b}{a}$ and $x_1\times x_2=\frac{c}{a}$

According to Ragini, she copied the constant term and the coefficient of $x^2$ correctly. Hence $3\times -\frac{1}{2}=-\frac{3}{2}=\frac{c}{a}$

And according to Gourav, he copied coefficient of $x$ and $x^2$ correctly. Hence $-1+(-\frac{1}{4})=-\frac{5}{4}=-\frac{b}{a}$

Hence the equation becomes,

$x^2+\frac{5}{4}x-\frac{3}{2}=0⟹4x^2+5x-6=0$

$⟹4x^2+8x-3x-6=0⟹4x(x+2)-3(x+2)=0⟹(4x-3)(x+2)=0⟹x_1=-2,x_2=\frac{3}{4}$

are the correct roots of the given equation.