For the equation : x^2+2x=4 find the exact roots.

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Cindy S. | A Heart for MathA Heart for Math

If you move the 4 over, x^2+2x-4=0, you get a quadratic that does not factor. So, you have two choices here. Use the quadratic formula or complete the square.

Quadratic formula: x = (-b +- sqrt(b^2 - 4ac))/(2a) Here, a=1, b=2, c=-4.

x= (-2 +- sqrt(4-4(1)(-4))/(2*1) = (-2 +- sqrt(20))/(2) now sqrt(20) = sqrt(4)*sqrt(5) =2 * sqrt(5)

so, x = (-2 +- 2*sqrt(5))/2 now factor out a 2: 2(-1 +- sqrt(5))/2 then cancel the 2's

x = -1 +- sqrt(5)

Completing the square: x^2 + 2x = 4 half of 2 is one, square one and you get one, that is what you add to both sides: x^2 + 2x + 1 = 4 + 1

(x + 1)^2 = 5

x + 1 = +- sqrt(5)

x = -1 +- sqrt(5)

And there's your answers.

Hey Melissa -- we may "re-frame" x^2 +2x -4 => (x^2 +2x +1) -5 => (x+1)(x+1) -5

... x+1 = +/- sqrt 5 ==> **x = -1 +/- sqrt 5 **... Best wishes, ma'am :)

First move everything to the same side of the equation

x^{2}+2x=4

x^{2}+2x-4=0

now we check the discriminate of the quadratic theorem to see what the roots are like:

b^{2}-4ac is 2^{2}-4(1)(-4) = 20

a positive discriminate indicates that there are two real roots to this equation. So finish the quadratic equation to find the roots.

-b+√20 = -2+√20

2a 2

and

-b+√20 = -2-√20

2a 2

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