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When solving the equation [tex]$2x^2 + 2x = 12$[/tex] by completing the square, which of the following represents the step before one would take the square root of both sides?

Select one:
a. [tex]$(x + \frac{1}{2})^2 = \frac{25}{4}$[/tex]
b. [tex][tex]$(x + \frac{1}{2})^2 = \frac{13}{2}$[/tex][/tex]
c. [tex]$(x + \frac{1}{4})^2 = \frac{25}{4}$[/tex]
d. [tex]$(2x + 1)^2 = 13$[/tex]


Sagot :

To solve the equation [tex]\(2x^2 + 2x = 12\)[/tex] by completing the square, we'll follow these steps:

1. Divide all terms by the coefficient of [tex]\(x^2\)[/tex]:
The equation is [tex]\(2x^2 + 2x = 12\)[/tex].

Divide by 2:
[tex]\[ x^2 + x = 6 \][/tex]

2. Move the constant term to the other side:
[tex]\[ x^2 + x - 6 = 0 \][/tex]
Add 6 to both sides:
[tex]\[ x^2 + x = 6 \][/tex]

3. Complete the square:
We need to turn the left side into a perfect square trinomial. To do this, take half the coefficient of [tex]\(x\)[/tex] (which is 1), square it, and add that square to both sides.

Half of 1 is [tex]\( \frac{1}{2} \)[/tex], and squaring it gives [tex]\( \left( \frac{1}{2} \right)^2 = \frac{1}{4} \)[/tex].

Add [tex]\( \frac{1}{4} \)[/tex] to both sides:
[tex]\[ x^2 + x + \frac{1}{4} = 6 + \frac{1}{4} \][/tex]
[tex]\[ x^2 + x + \frac{1}{4} = \frac{24}{4} + \frac{1}{4} \][/tex]
[tex]\[ x^2 + x + \frac{1}{4} = \frac{25}{4} \][/tex]

4. Write the left side as a square of a binomial:
[tex]\[ \left( x + \frac{1}{2} \right)^2 = \frac{25}{4} \][/tex]

Thus, the step before taking the square root of both sides is:
[tex]\[ (x+ \frac{1}{2})^2 = \frac{25}{4} \][/tex]

Therefore, the correct answer is:
a. [tex]\( (x + \frac{1}{2})^2 = \frac{25}{4} \)[/tex]