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ludmilkaskok [199]
3 years ago
12

2. Solve the equation using the quadratic formula. -4x2 + 40x + 16 = 0 Page 1 of 2

Mathematics
1 answer:
kumpel [21]3 years ago
8 0

The solution to given quadratic equation is x = -0.3851 or x = 10.3851

<em><u>Solution:</u></em>

Given equation is:

-4x^2 + 40x + 16 = 0

To find: solve by quadratic equation formula

\text {For a quadratic equation } a x^{2}+b x+c=0, \text { where } a \neq 0\\\\x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}

Using the above formula,

\text{ For} -4x^2 + 40x + 16 = 0 \text{ we have } a = -4 ; b = 40 ; c = 16

<u><em>Substituting the values of a = -4 ; b = 40 ; c = 16 in above quadratic formula we get,</em></u>

\begin{aligned}&x=\frac{-40 \pm \sqrt{40^{2}-4(-4)(16)}}{2 \times-4}\\\\&x=\frac{-40 \pm \sqrt{1600+256}}{-8}\\\\&x=\frac{-40 \pm 43.081}{-8}\\\\&x=\frac{-40+43.081}{-8} \text { or } x=\frac{-40-43.081}{-8}\\\\&x=-0.3851 \text { or } x=10.3851\end{aligned}

Thus the solution to given quadratic equation is x = - 0.3851 or x = 10.3851

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Answer: The log simplifies to -2

-------------------------------------------
-------------------------------------------

Explanation:

We will use the log rule that log(x^y) = y*log(x). Call this log rule 1. This log rule basically allows us to pull the exponent down.

Another log rule that we will use is \log_x\left(x\right) = 1 where x is any positive real number but x = 1 is NOT allowed. Call this log rule 2.

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So,

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_______________________________________

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