Question

If $\alpha ,\beta$ are the roots of a quadratic equation $x^2-3kx+k^2=0$, find the values of k. If ${\alpha }^2+{\beta }^2=\frac{7}{4}$.

• A. $\pm \frac{1}{3}$
• B. $\pm \frac{1}{2}$
• C. $\pm \frac{1}{4}$
• D. $\pm \frac{1}{5}$

Answer 1

The correct option is B $\pm \frac{1}{2}$

The given quadratic equation is $x^2-3kx+k^2=0$

Hence, $a=1,b=-3k,c=k^3$

$\because \alpha ,\beta$ are the roots of the given quadratic equation :

$\therefore \alpha +\beta =-\frac{b}{a}=-\frac{-3k}{1}=3k$

$\alpha .\beta =\frac{c}{a}=\frac{k^2}{1}=k^2$

$\therefore {(\alpha +\beta )}^2={\alpha }^2+{\beta }^2+2\alpha \beta$

$\Rightarrow {\left(3k\right)}^2=\frac{7}{4}+2k^2[\because {\alpha }^2+{\beta }^2=\frac{7}{4}]$

$\Rightarrow {9k}^2-2k^2=\frac{7}{4}$

$\Rightarrow {7k}^2=\frac{7}{4}\Rightarrow k^2=\frac{1}{4}$

$\therefore k\pm \sqrt{\frac{1}{4}}=\pm \frac{1}{2}$