Question

Find the value of $k$ for the following quadratic equation, so that they have two real and equal roots:

$k{x}^2-2\sqrt{5}x+4=0$

• A. $k=\frac{4}{5}$
• B. $k=-\frac{4}{5}$
• C. $k=\frac{5}{4}$
• D. $k=-\frac{5}{4}$

Answer 1

The correct option is C $k=\frac{5}{4}$

In $a{x}^2+bx+c=0,$
Discriminant $D=b^2-4ac$,

If $D>0$, the roots are real and distinct
If $D=0$, the roots are real and equal

In $K{x}^2-2\sqrt{5}x+4=0$

$a=k$, $b=-2\sqrt{5}$ and $c=4$

then $D=b^2-4ac=0$

Given roots are real and equal

So, $(-2\sqrt{5}{)}^2-4\times k\times 4=0$

$\Rightarrow 20-16k=0$

$\Rightarrow k=\frac{5}{4}$