Completing the squareIf we try to solve this quadratic equation by factoring,x2 + 6x + 2=0

we cannot. Therefore, we will complete the square. We will make the quadratic into the form

a2 + 2ab + b2 =(a + b)2 .

This technique is valid only when the coefficient of x2 is 1.1) Transpose the constant term to the rightx2 + 6x=2.2) Add a square number to both sides -- add the square of half the coefficient of x. In this case, add the square of half of 6; that is, add the square of 3, which is 9:

x2 + 6x + 9=2 + 9.The left-hand side is now the perfect square of (x + 3). (x + 3)2 = 7.3 is half of the coefficient 6. That equation has the forma2 =b

which implies

a=.

Therefore,

x + 3=

x=3 .

That is, the solutions tox2 + 6x + 2 = 0are the conjugate pair,3 + , 3 .For a method of checking these roots, see the theorem of the sum and product of the roots: Lesson 10 of Topics in Precalculus,In Lesson 18 there are examples and problems in which the coefficient of x is odd. Also, some of the quadratics below have complex roots, and some involve simplifying radicals.Problem 6.Solve each quadratic equation by completing the square.To see the answer, pass your cursor from left to rightover the colored area. To cover the answer again, click "Refresh" ("Reload").Do the problem yourself first! a) x2 2x 2=0b) x2 10x + 20=0

x2 2x =2x2 10x =20

x2 2x + 1 =2 + 1x2 10x + 25=20 + 25

(x 1)2 =3(x 5)2 =5

x 1=x 5=

x=1 x=5

c) x2 4x + 13=0d) x2 + 6x + 29=0

x2 4x =13x2 + 6x =29

x2 4x + 4 =13 + 4x2 + 6x + 9=29+ 9

(x 2)2 =9(x + 3)2 =20

x 2=3ix + 3=

x=2 3ix=3 2i

e) x2 5x 5=0f) x2 + 3x + 1=0

x2 5x =5x2 + 3x =1

x2 5x + 25/4=5 + 25/4x2 + 3x +9/4=1 +9/4

(x 5/2)2 =5 + 25/4(x + 3/2)2 = 1 + 9/4

x 5/2=x + 3/2=

x=x=

Proof of the quadratic formulaTo prove the quadratic formula, we complete the square. But to do that, the coefficient of x2 must be 1. Therefore, we will divide both sides of the original equation by a:

on multiplying both c and a by 4a, thus making the denominators the same

This is the quadratic formula.