Question

If $\alpha$ and $\beta$ are the roots of the equation $2x^2-3x-5=0$, then form a quadratic equation whose roots are ${\alpha }^2$ and ${\beta }^2$.

• A. $4x^2-29x+25=0$
• B. $4x^2+29+25=0$
• C. $4x^2-29-25=0$
• D. None of these

Answer 1

The correct option is A $4x^2-29x+25=0$

$\Rightarrow$ $\alpha$ and $\beta$ are two roots of the given quadratic equation $2x^2-3x-5=0$

$\Rightarrow$ $2x^2-3x-5=0$

$\Rightarrow$ $2x^2-5x+2x-5=0$

$\Rightarrow$ $x(2x-5)+1(2x-5)=0$

$\Rightarrow$ $(2x-5)(x+1)=0$

$\Rightarrow$ $2x-5=0$ and $x+1=0$

$\therefore$ $x=\frac{5}{2}$ and $x=-1$

$\therefore$ $\alpha =\frac{5}{2}$ and $\beta =-1$

$\Rightarrow$ ${\alpha }^2={\left(\frac{5}{2}\right)}^2=\frac{25}{4}$

$\Rightarrow$ ${\beta }^2=(-1{)}^2=1$

$\therefore$ New roots are $\frac{25}{4}$ and $1$.

New quadratic equation,

$\Rightarrow$ $(x-\frac{25}{4})(x-1)=0$

$\Rightarrow$ $x^2-x-\frac{25x}{4}+\frac{25}{4}=0$

$\Rightarrow$ $4x^2-4x-25x+25=0$

$\Rightarrow$ $4x^2-29x+25=0$