Some quadratic equations have no $x$ term, like $x^2 = 25$ or $(x - 3)^2 = 7$. Factoring may not be the easiest way; instead, undo the square with a square root. You will need simplified square roots, and the same reasoning about two solutions as in the zero product property.
Why it works
Which numbers square to $25$? Both $5$ and $-5$. The symbol $\sqrt{25}$ means only the positive one, $5$, so to get every solution we write $x = \pm\sqrt{25} = \pm 5$. Factoring agrees: $x^2 - 25 = (x - 5)(x + 5) = 0$. And if a square must equal a negative number, there is no real solution, since no real number squares to a negative.
Key idea
Isolate the square, then take the square root of both sides with $\pm$: if $u^2 = k$ with $k > 0$, then $u = \pm\sqrt{k}$. If $k = 0$ there is one solution; if $k < 0$, none.
Worked examples
Worked example: A shifted square
Solve $(x - 3)^2 = 7$.
- $x - 3 = \pm\sqrt{7}$.
- Add $3$: $x = 3 \pm \sqrt{7}$.
Answer: $x = 3 + \sqrt{7}$ or $x = 3 - \sqrt{7}$ (about $5.65$ and $0.35$).
Worked example: Isolate first
Solve $2(x + 1)^2 - 5 = 31$.
- Add $5$: $2(x + 1)^2 = 36$. Divide by $2$: $(x + 1)^2 = 18$.
- $x + 1 = \pm\sqrt{18} = \pm 3\sqrt{2}$.
- $x = -1 \pm 3\sqrt{2}$.
Answer: $x = -1 + 3\sqrt{2}$ or $x = -1 - 3\sqrt{2}$.
Common mistake
Keeping only the positive root of $x^2 = 25$, or taking the root before isolating the square. In $2(x + 1)^2 - 5 = 31$, you cannot take a square root of the left side until the square stands alone. Check: $(-5)^2 = 25$ too, so $-5$ must be in your answer.