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2 Help
3x² +8x+ 5-14
Which constant would you add to the RIGHT side of the
quadratic equation in order to solve by completing the
square? Remember, this is a two-step process​

2 Help3x 8x 514Which constant would you add to the RIGHT side of thequadratic equation in order to solve by completing thesquare Remember this is a twostep proc class=

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ANSWER

The correct answer is D

EXPLANATION

The given quadratic expression is

[tex]3 {x}^{2} + 8x = - 1[/tex]

Divide through by 3

[tex]{x}^{2} + \frac{8}{3} x = - \frac{1}{3} [/tex]

We now add the square of half the coefficient of x because the coefficient of the quadratic term is now unity.

The constant we would add is

[tex] {( \frac{4}{3} )}^{2} = \frac{16}{9} [/tex]

The correct answer is D

luisejr77

Answer: Option D.

Step-by-step explanation:

Given the quadratic equation [tex]3x^2+8x=-1[/tex], you can observe that the leading coefficient is not 1, then you have to divide both sides of the equation by 3:

[tex]\frac{3x^2+8x}{3}=\frac{-1}{3}\\\\x^2+\frac{8}{3}x=\frac{-1}{3}[/tex]

Now, you can observe that the coefficient of the x-term is [tex]\frac{3}{8}[/tex], so you must divide it by 2 and square it. Then you get:

[tex](\frac{\frac{8}{3}}{\frac{2}{1}})^2=\frac{16}{9}[/tex]

Therefore, you must add [tex]\frac{16}{9}[/tex] to both sides of the equation.

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