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 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