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Using the Completing the Square method, what are the zeros of the quadratic function [tex]f(x) = 2x^2 + 8x - 3[/tex]?

A. [tex]x = -2 - \sqrt{\frac{11}{2}}[/tex] and [tex]x = -2 + \sqrt{\frac{11}{2}}[/tex]
B. [tex]x = -2 - \sqrt{\frac{7}{2}}[/tex] and [tex]x = -2 + \sqrt{\frac{7}{2}}[/tex]
C. [tex]x = 2 - \sqrt{\frac{11}{2}}[/tex] and [tex]x = 2 + \sqrt{\frac{11}{2}}[/tex]
D. [tex]x = 2 - \sqrt{\frac{7}{2}}[/tex] and [tex]x = 2 + \sqrt{\frac{7}{2}}[/tex]


Sagot :

Let's determine the zeros of the quadratic function [tex]\( f(x) = 2x^2 + 8x - 3 \)[/tex] using the completing the square method.

1. Start with the given quadratic function:
[tex]\[ f(x) = 2x^2 + 8x - 3 \][/tex]

2. Rewrite it in the standard quadratic form:
[tex]\[ 2x^2 + 8x - 3 = 0 \][/tex]

3. Divide all terms by the leading coefficient (2) to simplify:
[tex]\[ x^2 + 4x - \frac{3}{2} = 0 \][/tex]

4. Move the constant term to the right side of the equation:
[tex]\[ x^2 + 4x = \frac{3}{2} \][/tex]

5. To complete the square, add and subtract the square of half the coefficient of [tex]\( x \)[/tex] (which is 4/2 = 2) inside the equation:
[tex]\[ x^2 + 4x + 4 - 4 = \frac{3}{2} \][/tex]

6. This can be rewritten as:
[tex]\[ (x + 2)^2 - 4 = \frac{3}{2} \][/tex]

7. Move the constant term (-4) to the right side:
[tex]\[ (x + 2)^2 = \frac{3}{2} + 4 \][/tex]
[tex]\[ (x + 2)^2 = \frac{3}{2} + \frac{8}{2} \][/tex]
[tex]\[ (x + 2)^2 = \frac{11}{2} \][/tex]

8. Take the square root of both sides:
[tex]\[ x + 2 = \pm\sqrt{\frac{11}{2}} \][/tex]

9. Solve for [tex]\( x \)[/tex]:
[tex]\[ x = -2 \pm \sqrt{\frac{11}{2}} \][/tex]

Thus, the solutions (zeros) to the equation [tex]\( 2x^2 + 8x - 3 = 0 \)[/tex] are:
[tex]\[ x = -2 - \sqrt{\frac{11}{2}} \][/tex]
[tex]\[ x = -2 + \sqrt{\frac{11}{2}} \][/tex]

So the correct answer is:
[tex]\[ \boxed{x = -2 - \sqrt{\frac{11}{2}} \text{ and } x = -2 + \sqrt{\frac{11}{2}}} \][/tex]