Which of the following is an extraneous solution of square root -3x-2=x+2

Answer:

A) -6

Explanation:

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The extraneous solution of the equation [tex]\(\sqrt{-3x - 2} = x + 2\) is \(x = -6\),[/tex] as it doesn't satisfy the original equation after substitution.

To find the extraneous solutions of the equation [tex]\(\sqrt{-3x - 2} = x + 2\),[/tex] we need to solve the equation and then check each solution to see if it satisfies the original equation.

Given equation: [tex]\(\sqrt{-3x - 2} = x + 2\)[/tex]

Step 1: Square both sides of the equation to eliminate the square root:

[tex](\sqrt{-3x - 2})^2 &= (x + 2)^2 \\[/tex]

[tex]-3x - 2 &= x^2 + 4x + 4[/tex]

Step 2: Rearrange the equation into a quadratic equation:

[tex]0 &= x^2 + 4x + 4 + 3x + 2 \\[/tex]

[tex]0 &= x^2 + 7x + 6[/tex]

Step 3: Solve the quadratic equation by factoring or using the quadratic formula.

The quadratic equation [tex]\(x^2 + 7x + 6 = 0\)[/tex] factors as [tex]\((x + 6)(x + 1) = 0\).[/tex]

So, the solutions are [tex]\(x = -6\) and \(x = -1\).[/tex]

Step 4: Check each solution to see if it satisfies the original equation:

For [tex]\(x = -6\):[/tex]

[tex]\sqrt{-3(-6) - 2} &= (-6) + 2 \\[/tex]

[tex]\sqrt{18 - 2} &= -4 \\[/tex]

[tex]\sqrt{16} &= -4 \\[/tex]

[tex]4 &= -4 \quad (\text{False})[/tex]

For [tex]\(x = -1\):[/tex]

[tex]\sqrt{-3(-1) - 2} &= (-1) + 2 \\[/tex]

[tex]\sqrt{3 - 2} &= 1 \\[/tex]

[tex]\sqrt{1} &= 1 \\[/tex]

[tex]1 &= 1 \quad (\text{True})[/tex]

Step 5: Therefore, the extraneous solution is [tex]\(x = -6\).[/tex]

The probable question may be:

Which of the following is an extraneous solution of [tex]\sqrt{-3x-2} = x+ 2[/tex] ?

x = -6

x = -1

x = 1

x = 6

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