Question

Avery solves the equation below by first squaring both sides of the equation.

√z^2+8 = 1 -2z
What extraneous solution does Avery obtain?
z= ?

Answer 1

`\left(\sqrt{z^(2)+8} \right)^(2)=(1-2z)^(2)\\\\z^(2)+8=4z^(2)-4z+1\\\\0=3z^(2)-4z-7\\\\0=(z+1)(3z-7)\\\\z=-1, z=(7)/(3)`

We know that if z=7/3, the right-hand side will be negative, which means it is an extraneous solution as square roots cannot be negative.