How many roots does -2x^2-8=0 have

Mathematics

1 answer:

Aloiza [94]3 years ago

4 0

Answer: two

Step-by-step explanation:

Factor out the equation:

-2( $x^{2}$ + 4) = 0

-2 cannot equal zero, so that means $x^{2}$ - 4 will have to equal to zero in order to get zero as a product

$x^{2}$ - 4 = 0

Factor the equation to get:

(x+2)(x-2) = 0

x = 2, x = -2

The equation has two roots

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Solve for x.

Daniel [21]

Answer:

$x=-\frac{9}{4} \\ or \:\\x = -2\frac{1}{4}$

Step-by-step explanation:

$-2 (x + 1/4)+1=5\\\\-2\left(x+\frac{1}{4}\right)+1=5\\\\\mathrm{Subtract\:}1\mathrm{\:from\:both\:sides}\\\\-2\left(x+\frac{1}{4}\right)+1-1=5-1\\\\Simplify\\-2\left(x+\frac{1}{4}\right)=4\\\\\mathrm{Divide\:both\:sides\:by\:}-2\\\frac{-2\left(x+\frac{1}{4}\right)}{-2}=\frac{4}{-2}\\\\Simplify\\x+\frac{1}{4}=-2\\\\\mathrm{Subtract\:}\frac{1}{4}\mathrm{\:from\:both\:sides}\\\\x+\frac{1}{4}-\frac{1}{4}=-2-\frac{1}{4}\\\\Simplify\\x=-\frac{9}{4}\\\\x = -2\frac{1}{4}$