**Completing the Square ex****amples:** Sometimes, the quadratic polynomials are not easily factorable.

**For Example**, consider x^2+4x-437=0

It is difficult to make factors of x^2+4x-437

In such a case the factorization and hence the solution of a quadratic equation can be found by the method of completing the square and extracting square roots.

Completing the Square examples

Say we have a simple expression like x^2 +bx. Having x twice in the same expression can make life hard. What can we do?

Well, with a little inspiration from Geometry we can convert it, like this:

**Example#1**

**Solve the equation x^2 + 4x - 437 = 0 by completing the square examples).**

**Solution:**

we have

x^2 + 4x - 437 = 0

x^2+2(x)(\frac42)=437

Now, for completing the square, adding both side (\frac42)^2

x^2+2(x)(\frac42)+(\frac42)^2=437+(\frac42)^2

⇒x^2+4x+(2)^2=437+(2)^2

⇒(x+2)^2=441

⇒(x+2)=\pm\sqrt{441}

⇒(x+2)=\pm21

⇒x=\pm21-2

⇒x=21-2 \;\;\;\;\;\;or\;\;\;\;\;\; x=-21-2

so\;\;\;\;\;\; x=19 \;\;\;\;\;\;and \;\;\;\;\;\;x=-23

Hence by completing the square

**Solution set={19,-23}**

Example#2

**Solve the equation x^2 -2x - 899 = 0 by completing the square examples.**

**Solution:**

we have

x^2 -2x - 899 = 0

x^2-2(x)(\frac22)=899

Now, for completing the square, adding both side (\frac22)^2

x^2+2(x)(\frac22)+(\frac22)^2=899+(\frac22)^2

⇒x^2-2x+(1)^2=899+(1)^2

⇒(x-1)^2=900

⇒(x-1)=\pm\sqrt{900}

⇒(x-1)=\pm30

⇒x=\pm30+1

⇒x=30+1 \;\;\;\;\;\;or\;\;\;\;\;\; x=-30+1

so\;\;\;\;\;\; x=31 \;\;\;\;\;\;and \;\;\;\;\;\;x=-29

Hence by completing the square

**Solution set={31.-29}**

you can also see quadratic formula