Question

which equation needs to be solved to find the possible solutions of $2x + \sqrt{x + 1} = 8$\
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a. $4x^2 - 33x + 63 = 0$\
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b. $4x^2 - 31x + 63 = 0$\
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c. $4x^2 - x + 63 = 0$\
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d. $4x^2 + x - 63 = 0$

Explanation:

Step1: Isolate the radical term

$\sqrt{x+1} = 8 - 2x$

Step2: Square both sides

$(\sqrt{x+1})^2 = (8 - 2x)^2$
$x + 1 = 64 - 32x + 4x^2$

Step3: Rearrange to standard quadratic

$4x^2 - 33x + 63 = 0$

Answer:

A. $4x^2 - 33x + 63 = 0$