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

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3 Answers

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
 
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