4.2 Solving by Square Roots

Notes

If the equation has only an $x^2$ term (or a squared binomial), isolate the square and take the square root of both sides.

• $x^2 = k$ gives $\displaystyle x = \pm\sqrt{k}$ when $k > 0$. If $k < 0$ there is no real solution.
• Example: $3x^2 - 48 = 0 \Rightarrow x^2 = 16 \Rightarrow x = \pm 4$.
• Example: $(x - 2)^2 = 25 \Rightarrow x - 2 = \pm 5 \Rightarrow x = 7$ or $x = -3$.
• Do not forget the $\pm$.