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Scrat [10]
3 years ago
10

How do you solve 8n squared+4n-16= -n squared using the quadratic formula?

Mathematics
1 answer:
klemol [59]3 years ago
3 0
8n² + 4n - 16 = -n²
8n² + n² + 4n - 16 = -n² + n²
9n² + 4n - 16 = 0
n = <u>-4 +/- √(4² - 4(9)(-16))</u>
                     2(9)
n = <u>-4 +/- √(16 + 576)</u>
                   18
n = <u>-4 +/- √(592)
</u>              18
n = <u>-4 +/- 24.33</u>
              18
n = -<u>4 + 24.33</u>        n = <u>-4 - 24.33</u>
             18                           18
n = <u>20.33</u>                n = <u>-28.33</u>
         18                            18
n = 1.1294              n = -1.57389
<u />
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Answer:

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Step-by-step explanation:

Consider the provided cubic function.

We need to find the equation having zeros: Square root of two, negative Square root of two, and -2.

A "zero" of a given function is an input value that produces an output of 0.

Substitute the value of zeros in the provided options to check.

Substitute x=-2 in f(x) = x^3 + 2x^2 - 2x + 4 .

f(x) = x^3 + 2x^2 - 2x + 4\\f(x) = (-2)^3 + 2(-2)^2 - 2(-2) + 4\\f(x) =-8 + 2(4)+4 + 4\\f(x) =8

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Substitute x=-2 in f(x) = x^3 + 2x^2 + 2x - 4 .

f(x) = x^3 + 2x^2 + 2x - 4\\f(x) = (-2)^3 + 2(-2)^2 + 2(-2) - 4\\f(x) =-8+2(4)-4-4\\f(x) =-8

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Substitute x=-2 in f(x) = x^3 - 2x^2 - 2x - 4 .

f(x) = x^3 - 2x^2 - 2x - 4\\f(x) = (-2)^3 - 2(-2)^2 - 2(-2) - 4\\f(x) =-8-8+4-4\\f(x) =-16

Therefore, the option is incorrect.

Substitute x=-2 in f(x) = x^3 + 2x^2 - 2x - 4 .

f(x) = x^3 + 2x^2 - 2x - 4\\f(x) = (-2)^3+2(-2)^2 - 2(-2) - 4\\f(x) =-8+8+4-4\\f(x) =0

Now check for other roots as well.

Substitute x=√2 in f(x) = x^3 + 2x^2 - 2x - 4 .

f(x) = x^3 + 2x^2 - 2x - 4\\f(x) = (\sqrt{2})^3+2(\sqrt{2})^2 - 2(\sqrt{2}) - 4\\f(x) =2\sqrt{2}+4-2\sqrt{2}-4\\f(x) =0

Substitute x=-√2 in f(x) = x^3 + 2x^2 - 2x - 4 .

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Therefore, the option is correct.

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