Question

In the following equation, determine whether the given values are solutions of the given :
$x^2-\sqrt{2x}-4=0,x=-\sqrt{2},x=-2\sqrt{2}$

• A. x = $-\sqrt{2}$ is not a solution but x = -2$\sqrt{2}$ is a solution.
• B. x = $-\sqrt{2}$ & x = -2$\sqrt{2}$ are not solutions.
• C. x = $-\sqrt{2}$ is a solution but x = -2$\sqrt{2}$ is not a solution.
• D. x = $-\sqrt{2}$ & x = -2$\sqrt{2}$ are solutions.

Answer 1

The correct option is C x = $-\sqrt{2}$ is a solution but x = -2$\sqrt{2}$ is not a solution.

Given equation is,

$x^2-\sqrt{2}x-4=0,x=-\sqrt{2},x=-2\sqrt{2}$

if these value of $x$ satisfy the given equation,

They must be the solutions,

Let $x=-\sqrt{2}$,

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

$2+2-4=0$

and $x=-2\sqrt{2}$

$(-2\sqrt{2})-\sqrt{2}.(-2\sqrt{2})-4$

$=8+4-4$

$=8\ne 0$

Hence $x=-\sqrt{2}$ is the solution but $x=-2\sqrt{2}$ is not.