The potential solutions to the radical equation sq root of a+5=a+3 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.

Option A. the solution a = -4 is an extraneous solution.

Step-by-step explanation:

Lets solve the radical equation to get the potential solutions.

√(a + 5) = (a + 3)

By squaring terms on both the sides

[√(a + 5)]² = (a +3)²

(a + 5) = (a + 3)²

(a + 5) = a² + 6a + 9

0 = a² + 6a + 9 - a - 5

a² + 5a + 4 = 0

a² + 4a + a + 4 + 0

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

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

a = -1 or a = -4

Now if we put the value of a = - 4 in the radical equation then we get

√(-4 + 5) = (-4 + 3)

√1 = -1

1 = -1 ( Therefore a = -4 is not the solution)

Now by putting a = -1

√(-1 +5) = -1 + 3

√4 = 2

2 = 2 which shows the equal relation.

Therefore a = -4 is an extraneous solution.

Option A. is the correct option.

it would open into a square depending on the type of envelope i’m not sure about the second part