Question

Solve the radical expression. Check your solution.
\[ \sqrt{2-x} =x\]

Answer 1

The given radical expression is $\sqrt{2-x}=x$

Squaring on both the sides of the equation, we get

$2-x=x^{2}$

Bringing all the variables and constant to the right side of the expression, so we get:

$x^{2}+x-2=0$

By comparing the above expression with the standard form $ax^{2}+bx+c=0$

we get a=1, b=1 and c= -2

Discrimant(D) = $b^{2}-4ac$

Discrimant(D)=$(1)^{2}-4(1)(-2)$

D=9

$x=\frac{-b+\sqrt{D}}{2a} and x=\frac{-b-\sqrt{D}}{2a}$

$x=\frac{-1+\sqrt{9}}{2} and x=\frac{-1-\sqrt{9}}{2}$

$x=\frac{-1+3}{2} and x=\frac{-1-3}{2}$

x= -1 and 2 are the required solutions of the given radical expression.