Solve the equation |x^2 - 13| = 3.

Step-by-step solution

1. 1

   Split on the definition of absolute value. $|A| = 3$ means the inside is $3$ units from zero in either direction, so $A = 3$ or $A = -3$. Here that gives two separate quadratic equations:

   $x^2 - 13 = 3 \qquad \text{or} \qquad x^2 - 13 = -3$

   Both branches are live because $3 > 0$; if the right-hand side were negative there would be no solutions at all.

2. 2

   Solve the positive branch.

   $x^2 - 13 = 3 \;\Longrightarrow\; x^2 = 16 \;\Longrightarrow\; x = \pm 4$

3. 3

   Solve the negative branch.

   $x^2 - 13 = -3 \;\Longrightarrow\; x^2 = 10 \;\Longrightarrow\; x = \pm\sqrt{10}$

   Both right-hand sides came out positive, so each branch really does contribute two real roots — the reason this equation has four solutions rather than two.

4. 4

   Verify every root. $|4^2 - 13| = |3| = 3$ and $|(-4)^2 - 13| = 3$. For the irrational pair, $(\pm\sqrt{10})^2 = 10$, so $|10 - 13| = |-3| = 3$. All four check exactly.

5. 5

   Read off the solution set. Ordered on the number line with $\sqrt{10} \approx 3.1623$:

   $x = -4,\ -\sqrt{10},\ \sqrt{10},\ 4$