Question

find the value (s) of k for which the quadratic equation
$x {}^{2} + 2 \sqrt{2kx} + 18 = 0$
has equal roots​

Answer 1

Answer:

k= 4.5/2

Step-by-step explanation:

For the quadratic equation to have real roots.

${b}^{2} - 4ac = 0$

${(2 \sqrt{2k} )}^{2} - 4 \times 18 = 0$

$8k = 18$

k = 4.5/2

Answer 2

Answer:

The value of k is

$\boxed{\red{\sf\:k\:=\:3}}\:\:\:\sf\:or\:\:\:\boxed{\red{\sf\:k\:=\:-\:3}}$

Step-by-step-explanation:

The given quadratic equation is $\sf\:x^{2}\:+\:2\:\sqrt{2k}\:x\:+\:18\:=\:0$

We have given that, the given quadratic equation has real and equal roots.

It means the value of discriminant must be 0.

$\therefore\sf\:b^{2}\:-\:4\:ac\:=\:0\\\\\implies\sf\:D\:=\:(\:2\:\sqrt{\cancel{2k}}\:)^{\cancel{2}}\:-\:4\:\times\:1\:\times\:18\\\\\implies\sf\:4\:\times\:2k^{2}\:-\:72\\\\\implies\sf\:8k^{2}\:-\:72\\\\\sf\:But,\:D\:=\:0\\\\\therefore\sf\:8k^{2}\:-\:72\:=\:0\\\\\implies\sf\:8k^{2}\:=\:72\\\\\implies\sf\:k^{2}\:=\:\frac{\cancel{72}}{\cancel8}\\\\\implies\sf\:k^{2}\:=\:9\\\\\implies\sf\:k\:=\:\pm\:3\\\\\implies\boxed{\red{\sf\:k\:=\:3}}\:\:\:\sf\:or\:\:\:\boxed{\red{\sf\:k\:=\:-\:3}}$