What value of [tex]\( g \)[/tex] makes the equation true?

[tex]\[
(x+7)(x-4) = x^2 + 9x - 29
\][/tex]

A. [tex]\(-11\)[/tex]

B. [tex]\(-3\)[/tex]

C. [tex]\(3\)[/tex]

D. [tex]\(11\)[/tex]



Answer :

To find the value of [tex]\( x \)[/tex] that makes the equation true, we will solve the equation step by step.

Given equation:
[tex]\[ (x + 7)(x - 4) = x^2 + 9x - 29 \][/tex]

Step 1: Expand the left side of the equation.

[tex]\[ (x + 7)(x - 4) \][/tex]

Expanding this using the distributive property (FOIL method), we get:

[tex]\[ x \cdot x + x \cdot (-4) + 7 \cdot x + 7 \cdot (-4) \][/tex]

[tex]\[ = x^2 - 4x + 7x - 28 \][/tex]

Combine like terms:

[tex]\[ x^2 + 3x - 28 \][/tex]

Now the equation looks like this:

[tex]\[ x^2 + 3x - 28 = x^2 + 9x - 29 \][/tex]

Step 2: Subtract [tex]\( x^2 + 9x - 29 \)[/tex] from both sides of the equation to set it to 0:

[tex]\[ (x^2 + 3x - 28) - (x^2 + 9x - 29) = 0 \][/tex]

Simplify the equation:

[tex]\[ x^2 + 3x - 28 - x^2 - 9x + 29 = 0 \][/tex]

Combine like terms:

[tex]\[ 3x - 9x - 28 + 29 = 0 \][/tex]

[tex]\[ -6x + 1 = 0 \][/tex]

Step 3: Solve for [tex]\( x \)[/tex]:

[tex]\[ -6x + 1 = 0 \][/tex]

Subtract 1 from both sides:

[tex]\[ -6x = -1 \][/tex]

Divide both sides by -6:

[tex]\[ x = \frac{1}{6} \][/tex]

Therefore, the value of [tex]\( g \)[/tex] that makes the equation true is:

[tex]\[ \frac{1}{6} \][/tex]

This value is not provided in the given options of -11, -3, 3, or 11. None of the provided options satisfy the equation, since the correct solution is [tex]\( \frac{1}{6} \)[/tex].

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