Step-by-Step Solution

Solve the radical equation: $\sqrt{2x + 8} + \sqrt{x + 5} = 7$

Step-by-Step Solution

Step 1: Isolate one radical and square both sides

$\sqrt{2x + 8} = 7 - \sqrt{x + 5}$

$\left(\sqrt{2x + 8}\right)^2 = \left(7 - \sqrt{x + 5}\right)^2$

$2x + 8 = 49 - 14\sqrt{x + 5} + (x + 5)$

$x - 46 = -14\sqrt{x + 5}$

Step 2: Square both sides again

$(x - 46)^2 = \left(-14\sqrt{x + 5}\right)^2$

$x^2 - 92x + 2116 = 196(x + 5)$

$x^2 - 92x + 2116 = 196x + 980$

$x^2 - 288x + 1136 = 0$

Step 3: Solve by quadratic formula

$x = \frac{288 \pm \sqrt{(-288)^2 - 4(1)(1136)}}{2} = \frac{288 \pm \sqrt{78400}}{2} = \frac{288 \pm 280}{2}$

$x = \frac{288 - 280}{2} = 4 \quad \text{or} \quad x = \frac{288 + 280}{2} = 284$

Step 4: Check for extraneous roots

• For $x = 4$: $\sqrt{2(4) + 8} + \sqrt{4 + 5} = \sqrt{16} + \sqrt{9} = 4 + 3 = 7 = \text{RHS} \quad \checkmark$

• For $x = 284$: $\sqrt{2(284) + 8} + \sqrt{284 + 5} = \sqrt{576} + \sqrt{289} = 24 + 17 = 41 \ne 7 \quad (\text{extraneous})$

$\boxed{\text{Solution Set: } \{4\}}$