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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.

Respuesta :

MrRoyal

Answer:

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

Step-by-step explanation:

Given

[tex]\sqrt{a + 5} = a + 3[/tex]

[tex]a = -4; a = -1[/tex]

Required

The true statement about the solutions

We have:

[tex]\sqrt{a + 5} = a + 3[/tex]

Square both sides

[tex]a + 5 = (a + 3)^2[/tex]

[tex]a + 5 =a^2 + 6a + 9[/tex]

Collect like terms

[tex]a^2 + 6a - a + 9 - 5 = 0[/tex]

[tex]a^2 + 5a + 4 = 0[/tex]

Expand

[tex]a^2 + 4a + a + 4 = 0[/tex]

Factorize

[tex]a(a + 4) + 1(a + 4) = 0[/tex]

Factor out a + 4

[tex](a + 1)(a + 4) = 0[/tex]

Split:

[tex]a + 1 = 0; a + 4 = 0[/tex]

Solve:

[tex]a =-1; a = -4[/tex]

Answer:

C on 3dge

Step-by-step explanation:

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

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