Question

The potential solutions to the radical equation are a = −4 and a = −1. Which statement is true about these solutions? The solution a = −4 is an extraneous solution. The solution a = −1 is an extraneous solution. Both a = −4 and a = −1 are true solutions. Neither a = −4 nor a = −1 are true solutions.

Answer 1

Answer:

Both a = −4 and a = −1 are true solutions

Step-by-step explanation:

Given

$\sqrt{a + 5} = a + 3$

$a = -4; a = -1$

Required

The true statement about the solutions

We have:

$\sqrt{a + 5} = a + 3$

Square both sides

$a + 5 = (a + 3)^2$

$a + 5 =a^2 + 6a + 9$

Collect like terms

$a^2 + 6a - a + 9 - 5 = 0$

$a^2 + 5a + 4 = 0$

Expand

$a^2 + 4a + a + 4 = 0$

Factorize

$a(a + 4) + 1(a + 4) = 0$

Factor out a + 4

$(a + 1)(a + 4) = 0$

Split:

$a + 1 = 0; a + 4 = 0$

Solve:

$a =-1; a = -4$