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3 years ago

What is the soulution of x^2 + 64 = 0

2 answers:

4 0

8i.

This boils down to what is the square root of -64. Negative numbers can't be squares so an imaginary number is part of your answer.

lys-0071 [83]3 years ago

3 0

Solve out
x^2 + 64 = 0
x^2 = -64 ( we are going to use imaginary numbers to solve)
x = $i \sqrt{64}$
x = 8i, -8i final answers

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x+y+2=-5 -2x-y+2=-1 x-2y-2=0 solve algebraically

x+y+2=-5 equation 1

-2x-y+2=-1 equation 2

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Add equation 1 + 2+ 3

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Find an equation for the line that passes through the points (5.-5) and (-4,-2)

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$\large\boxed{y=-\dfrac{1}{3}x-\dfrac{10}{3}\to x+3y=-10}$

Step-by-step explanation:

The slope-intercept form of an equation of a line:

$y=mx+b$

m - slope

b - y-intercept

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$m=\dfrac{y_2-y_1}{x_2-x_1}$

We have the points (5, -5) and (-4, -2).

Substiute:

$m=\dfrac{-2-(-5)}{-4-5}=\dfrac{-2+5}{-9}=\dfrac{3}{-9}=-\dfrac{1}{3}$

Put the value of the slope and the coordinates of the point (5, -5) to the equation of a line:

$-5=-\dfrac{1}{3}(5)+b$

$-5=-\dfrac{5}{3}+b$ add 5/3 to both sides

$-\dfrac{15}{3}+\dfrac{5}{3}=b\\\\-\dfrac{10}{3}=b\to b=-\dfrac{10}{3}$

Finally we have the equation of a line in the slope-intercept form:

$y=-\dfrac{1}{3}x-\dfrac{10}{3}$

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