Completing the Square examples: 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
