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

Question

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

Answer 1

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.

Answer 2

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 :)

Answer 3

First move everything to the same side of the equation

 

x²+2x=4

 

x²+2x-4=0

 

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

 

b²-4ac is 2²-4(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