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To solve the equation [tex]\(\sqrt{-2x - 5} - 4 = x\)[/tex], follow these steps:
1. Isolate the square root term: Start by isolating the square root expression.
[tex]\[\sqrt{-2x - 5} = x + 4\][/tex]
2. Square both sides: To eliminate the square root, square both sides of the equation.
[tex]\[(\sqrt{-2x - 5})^2 = (x + 4)^2\][/tex]
3. Simplify:
[tex]\[-2x - 5 = (x + 4)^2\][/tex]
Expand the right-hand side:
[tex]\[-2x - 5 = x^2 + 8x + 16\][/tex]
4. Form a quadratic equation: Bring all terms to one side of the equation to form a quadratic equation.
[tex]\[0 = x^2 + 8x + 16 + 2x + 5\][/tex]
Combine like terms:
[tex]\[x^2 + 10x + 21 = 0\][/tex]
5. Solve the quadratic equation: Use the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = 10\)[/tex], and [tex]\(c = 21\)[/tex].
[tex]\[x = \frac{-10 \pm \sqrt{10^2 - 4 \cdot 1 \cdot 21}}{2 \cdot 1}\][/tex]
[tex]\[x = \frac{-10 \pm \sqrt{100 - 84}}{2}\][/tex]
[tex]\[x = \frac{-10 \pm \sqrt{16}}{2}\][/tex]
[tex]\[x = \frac{-10 \pm 4}{2}\][/tex]
[tex]\[x = \frac{-10 + 4}{2} \quad \text{and} \quad x = \frac{-10 - 4}{2}\][/tex]
Simplify:
[tex]\[x = \frac{-6}{2} = -3 \quad \text{and} \quad x = \frac{-14}{2} = -7\][/tex]
6. Check the solutions: Substitute both [tex]\(x = -3\)[/tex] and [tex]\(x = -7\)[/tex] back into the original equation to check for extraneous solutions.
For [tex]\(x = -3\)[/tex]:
[tex]\[\sqrt{-2(-3) - 5} - 4 = -3\][/tex]
[tex]\[\sqrt{6 - 5} - 4 = -3\][/tex]
[tex]\[\sqrt{1} - 4 = -3\][/tex]
[tex]\[1 - 4 = -3\][/tex]
This holds true.
For [tex]\(x = -7\)[/tex]:
[tex]\[\sqrt{-2(-7) - 5} - 4 = -7\][/tex]
[tex]\[\sqrt{14 - 5} - 4 = -7\][/tex]
[tex]\[\sqrt{9} - 4 = -7\][/tex]
[tex]\[3 - 4 = -1\][/tex]
This does not hold true.
So, the correct solution is [tex]\(x = -3\)[/tex].
Therefore, the answer is:
C. [tex]\(\quad -3\)[/tex]
1. Isolate the square root term: Start by isolating the square root expression.
[tex]\[\sqrt{-2x - 5} = x + 4\][/tex]
2. Square both sides: To eliminate the square root, square both sides of the equation.
[tex]\[(\sqrt{-2x - 5})^2 = (x + 4)^2\][/tex]
3. Simplify:
[tex]\[-2x - 5 = (x + 4)^2\][/tex]
Expand the right-hand side:
[tex]\[-2x - 5 = x^2 + 8x + 16\][/tex]
4. Form a quadratic equation: Bring all terms to one side of the equation to form a quadratic equation.
[tex]\[0 = x^2 + 8x + 16 + 2x + 5\][/tex]
Combine like terms:
[tex]\[x^2 + 10x + 21 = 0\][/tex]
5. Solve the quadratic equation: Use the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = 10\)[/tex], and [tex]\(c = 21\)[/tex].
[tex]\[x = \frac{-10 \pm \sqrt{10^2 - 4 \cdot 1 \cdot 21}}{2 \cdot 1}\][/tex]
[tex]\[x = \frac{-10 \pm \sqrt{100 - 84}}{2}\][/tex]
[tex]\[x = \frac{-10 \pm \sqrt{16}}{2}\][/tex]
[tex]\[x = \frac{-10 \pm 4}{2}\][/tex]
[tex]\[x = \frac{-10 + 4}{2} \quad \text{and} \quad x = \frac{-10 - 4}{2}\][/tex]
Simplify:
[tex]\[x = \frac{-6}{2} = -3 \quad \text{and} \quad x = \frac{-14}{2} = -7\][/tex]
6. Check the solutions: Substitute both [tex]\(x = -3\)[/tex] and [tex]\(x = -7\)[/tex] back into the original equation to check for extraneous solutions.
For [tex]\(x = -3\)[/tex]:
[tex]\[\sqrt{-2(-3) - 5} - 4 = -3\][/tex]
[tex]\[\sqrt{6 - 5} - 4 = -3\][/tex]
[tex]\[\sqrt{1} - 4 = -3\][/tex]
[tex]\[1 - 4 = -3\][/tex]
This holds true.
For [tex]\(x = -7\)[/tex]:
[tex]\[\sqrt{-2(-7) - 5} - 4 = -7\][/tex]
[tex]\[\sqrt{14 - 5} - 4 = -7\][/tex]
[tex]\[\sqrt{9} - 4 = -7\][/tex]
[tex]\[3 - 4 = -1\][/tex]
This does not hold true.
So, the correct solution is [tex]\(x = -3\)[/tex].
Therefore, the answer is:
C. [tex]\(\quad -3\)[/tex]
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