Now use square roots and cube roots to solve equations. The sign rules from multiplying integers decide how many solutions there are.
Why it works
Both $5^{2} = 25$ and $(-5)^{2} = 25$, because a negative times a negative is positive. So $x^{2} = 25$ has two solutions. And no number squared is negative, so $x^{2} = -9$ has none.
Cubing keeps the sign: $2^{3} = 8$ but $(-2)^{3} = -8$. Each number has exactly one cube root, so $x^{3} = -8$ has exactly one solution.
Cubing keeps the sign: $2^{3} = 8$ but $(-2)^{3} = -8$. Each number has exactly one cube root, so $x^{3} = -8$ has exactly one solution.
Key idea
$x^{2} = p$ has two solutions, $x = \pm\sqrt{p}$, when $p > 0$; one solution, $x = 0$, when $p = 0$; and no solution when $p < 0$. $x^{3} = p$ always has exactly one solution, $x = \sqrt[3]{p}$.
Worked examples
Worked example: Two solutions
Solve $x^{2} = 49$.
- $7^{2} = 49$ and $(-7)^{2} = 49$.
- Check both in the equation: each gives 49.
Answer: $x = 7$ or $x = -7$, written $x = \pm 7$
Worked example: Get the square alone first
Solve $3x^{2} = 48$.
- Divide both sides by 3: $x^{2} = 16$.
- Take square roots: $x = \pm 4$.
Answer: $x = \pm 4$
Worked example: A root that isn’t whole
Solve $x^{2} = 10$.
- 10 is not a perfect square, so leave the answer as a root.
- $x = \sqrt{10}$ or $x = -\sqrt{10}$ (about $\pm 3.16$).
Answer: $x = \pm\sqrt{10}$
Worked example: A cube
Solve $x^{3} = -125$.
- $(-5)^{3} = -125$.
- $5^{3} = 125$ is positive, so it does not work. There is just one solution.
Answer: $x = -5$
Common mistake
Giving only $x = 5$ for $x^{2} = 25$. Check $-5$ too: $(-5)^{2} = 25$, so it is also a solution. And watch for negatives on the right: $x^{2} = -9$ has no solution (not $\pm 3$, because $3^{2} = 9$).