Equations involving squaring a variable under a radical sign

Question

If anyone can help me with how to go about solving these kind of equations i would really appreciate it. :-)

$$\sqrt{36-2x^2} = 4$$

Solve for X

Answer 1

$36-2x^2 = 16$

$\frac{36-16}{2} = x^2$

$x= \pm \sqrt{10}.$

Answer 2

I would start by looking at how to get rid of the square root sign on the left-hand side of the equation. Then I would look at isolating your variable, i.e. x. What did you attempt already?

Answer 3

Square both sides:

$$36-2x^2=16$$ $$ -2x^2+20=0$$ $$ 2x^2-20=0$$ $$x^2-10=0$$ $$x = ±\sqrt{10}$$

Now we need to check for extraneous solutions. You can then substitute these values of $x$ in the LHS, and check they equal the RHS.