Question

Find the value of k for which quadratic equation root 2x square - kx + root 2 = 0 Has two equal roots

Answer 1

For the equation ,$ax {}^{2} + bx + c = 0$, has 2 equal root then discriminant , d = 0

$d = {b}^{2} - 4ac$

Here the equation is $2x { }^{2} - kx + \sqrt{2} = 0$, then its discriminant is

$d = { (- k)}^{2} - 4 \times 2 \times \sqrt{2} = 0$

then,

${k}^{2} = 8 \sqrt{2} \\ \\ k = \sqrt{8 \sqrt{2} } = 2 \sqrt{2} \sqrt[4]{2}$

Answer 2

Answer:

$k=2\sqrt{2}$

Step-by-step explanation:

Let a quadratic equation is defined as

$ax^2+bx+c=0$

If $b^2-4ac=0$, then the above equation has two equal roots.

The given quadratic equation is

$\sqrt{2}x^2-kx+\sqrt{2}=0$

Here, $a=\sqrt{2},b=-k,c=\sqrt{2}$

It is given that the given equation has 2 equal roots.

$b^2-4ac=0$

$(-k)^2-4(\sqrt{2})(\sqrt{2})=0$

$k^2-4(2)=0$

$k^2-8=0$

Add 8 on both sides.

$k^2=8$

Taking square root on both sides.

$k=\sqrt{8}$

$k=2\sqrt{2}$

Therefore the value of k is $k=2\sqrt{2}$.