Completing the square for a quadratic in standard form, x2 + bx +c requires writing it in vertex form

( x-h )2 + k.

You are watching: Solve x2 + 6x + 4 = 0.

The choice of h and k depends on factoring the original trinomial as a perfect square.

1.The first step is to set the expression to zero

x2+6x+4=0

2.The second step is to add constants to both sides to make the expression a perfect square trinomial.

In this case, adding 5 to both sides makes it a perfect square because:

x2+2(3x) + ( 3)2 =0 + 5 or x2+ 2(3x) + 4 + 5 = 0+ 5

3.We have now used the mathematical identity: ( x+ a) 2 = x2 + 2ax + a2 to simplify our problem

4.The problem to solve becomes : x2+2(3x) + ( 3)2 = (x + 3) 2= 5

5.Taking the square root of both sides: sqrt((x+3)2 ) = +/- sqrt(5) or x+3 = +/- sqrt(5)

6.Solving for x: x = -3 - sqrt(5) and x = -3 + sqrt(5)


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Madina A.

h = to what ??k =5
Reharbor

Gladys S.

so in vertex form the equation is (x--3)^2 +5. Is a upward parabola centered at x = -3 and is shifted vertically by 5 .
Reharbor

Gladys S.

h =-3; k = 5
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Gladys S.

standard form is (x-h)^2 + k in this case : h = -3 and k = 5
Reharbor

Gladys S.

standard form is ( x -- 3)^2 + -5 . h = -3 and k = -5 .upward parabola centered at 0. passing thru x = -3 - sqrt(5) and -3+sqrt(5). Minimum at x = 0. where y = -5.
Reharbor

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