Question

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

$2x^2-(k-2)x+1=0$

• A. $k=-2\pm \sqrt{2}$
• B. $k=-2\pm 2\sqrt{2}$
• C. $k=2\pm \sqrt{2}$
• D. $k=2\pm 2\sqrt{2}$

Answer 1

The correct option is D $k=2\pm 2\sqrt{2}$

Given Quadratic equation is $2x^2-(k-2)x+1=0$

Now, given that the roots are equal and real.

For a quadratic equation $a{x}^2+bx+c=0$, roots are real and equal if $D=b^2-4ac=0$.

Here, $a=2,b=-(k-2),c=1$

Therefore, $D=b^2-4ac=0$

$\Rightarrow (-(k-2){)}^2-4\times 2\times 1=0$
$\Rightarrow k^2-2k\times 2+4-8=0$
$\Rightarrow k^2-4k-4=0$

This is a quadratic equation whose roots are

$\Rightarrow k=\frac{-(-4)\pm \sqrt{(-4{)}^2-4\times 1\times (-4)}}{2\times 1}$

$\Rightarrow k=\frac{4\pm \sqrt{16+16}}{2}$

$\Rightarrow k=\frac{4\pm \sqrt{32}}{2}$

$\Rightarrow k=\frac{4\pm 4\sqrt{2}}{2}$

$\Rightarrow k=\frac{4(1\pm \sqrt{2})}{2}$

$\Rightarrow k=2\pm 2\sqrt{2}$