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3 years ago

Complete the square for each expression. Then factor the Trinomial x^2+8x

Mathematics

1 answer:

Goryan [66]3 years ago

8 0

The value of x is $x=0$ or $x=-8$

Step-by-step explanation:

The expression is $x^{2} +8x=0$

To complete the square, the equation is of the form $ax^{2} +bx+c=0$

The constant term c can be determined using, $c=\left(\frac{\frac{b}{a}}{2}\right)^{2}$

$\begin{aligned}c &=\left(\frac{8}{2}\right)^{2} \\&=\left(\frac{8}{2}\right)^{2} \\c &=4^{2} \\c &=16\end{aligned}$

Rewriting the expression $x^{2} +8x=0$ and factoring the trinomial, we have,

$\begin{array}{r}{x^{2}+8 x+16=16} \\{(x+4)^{2}=16}\end{array}$

Taking square root on both sides, we get,

$\begin{aligned}&x+4=\sqrt{16}\\&x+4=\pm 4\end{aligned}$

Either,

$\begin{array}{r}{x+4=4} \\{x=0}\end{array}$ or $\begin{array}{r}{x+4=-4} \\{x=-8}\end{array}$

Thus, the value of x is $x=0$ or $x=-8$

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3 years ago

Calculus Problem

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The two parabolas intersect for

$8-x^2 = x^2 \implies 2x^2 = 8 \implies x^2 = 4 \implies x=\pm2$

and so the base of each solid is the set

$B = \left\{(x,y) \,:\, -2\le x\le2 \text{ and } x^2 \le y \le 8-x^2\right\}$

The side length of each cross section that coincides with B is equal to the vertical distance between the two parabolas, $|x^2-(8-x^2)| = 2|x^2-4|$ . But since -2 ≤ x ≤ 2, this reduces to $2(x^2-4)$ .

a. Square cross sections will contribute a volume of

$\left(2(x^2-4)\right)^2 \, \Delta x = 4(x^2-4)^2 \, \Delta x$

where ∆x is the thickness of the section. Then the volume would be

$\displaystyle \int_{-2}^2 4(x^2-4)^2 \, dx = 8 \int_0^2 (x^2-4)^2 \, dx \\\\ = 8 \int_0^2 (x^4-8x^2+16) \, dx \\\\ = 8 \left(\frac{2^5}5 - \frac{8\times2^3}3 + 16\times2\right) = \boxed{\frac{2048}{15}}$

where we take advantage of symmetry in the first line.

b. For a semicircle, the side length we found earlier corresponds to diameter. Each semicircular cross section will contribute a volume of

$\dfrac\pi8 \left(2(x^2-4)\right)^2 \, \Delta x = \dfrac\pi2 (x^2-4)^2 \, \Delta x$

We end up with the same integral as before except for the leading constant:

$\displaystyle \int_{-2}^2 \frac\pi2 (x^2-4)^2 \, dx = \pi \int_0^2 (x^2-4)^2 \, dx$

Using the result of part (a), the volume is

$\displaystyle \frac\pi8 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{256\pi}{15}}}$

c. An equilateral triangle with side length s has area √3/4 s², hence the volume of a given section is

$\dfrac{\sqrt3}4 \left(2(x^2-4)\right)^2 \, \Delta x = \sqrt3 (x^2-4)^2 \, \Delta x$

and using the result of part (a) again, the volume is

$\displaystyle \int_{-2}^2 \sqrt 3(x^2-4)^2 \, dx = \frac{\sqrt3}4 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{512}{5\sqrt3}}$

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2 years ago

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