Completing the Square examples: Sometimes, the quadratic polynomials are not easily factorable.
For Example, consider [katex]x^2+4x-437=0[/katex]
It is difficult to make factors of [katex]x^2+4x-437[/katex]
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 [katex]x^2 +bx[/katex]. Having [katex]x [/katex] 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 [katex]x^2 + 4x – 437 = 0[/katex] by completing the square examples).
Solution:
we have
[katex]x^2 + 4x – 437 = 0[/katex]
[katex]x^2+2(x)(\frac42)=437[/katex]
Now, for completing the square, adding both side [katex](\frac42)^2[/katex]
[katex]x^2+2(x)(\frac42)+(\frac42)^2=437+(\frac42)^2[/katex]
[katex]⇒x^2+4x+(2)^2=437+(2)^2[/katex]
[katex]⇒(x+2)^2=441[/katex]
[katex]⇒(x+2)=\pm\sqrt{441}[/katex]
[katex]⇒(x+2)=\pm21[/katex]
[katex]⇒x=\pm21-2[/katex]
[katex]⇒x=21-2 \;\;\;\;\;\;or\;\;\;\;\;\; x=-21-2[/katex]
[katex]so\;\;\;\;\;\; x=19 \;\;\;\;\;\;and \;\;\;\;\;\;x=-23[/katex]
Hence by completing the square
Solution set={19,-23}
Example#2
Solve the equation [katex]x^2 -2x – 899 = 0[/katex] by completing the square examples.
Solution:
we have
[katex]x^2 -2x – 899 = 0[/katex]
[katex]x^2-2(x)(\frac22)=899[/katex]
Now, for completing the square, adding both side [katex](\frac22)^2[/katex]
[katex]x^2+2(x)(\frac22)+(\frac22)^2=899+(\frac22)^2[/katex]
[katex]⇒x^2-2x+(1)^2=899+(1)^2[/katex]
[katex]⇒(x-1)^2=900[/katex]
[katex]⇒(x-1)=\pm\sqrt{900}[/katex]
[katex]⇒(x-1)=\pm30[/katex]
[katex]⇒x=\pm30+1[/katex]
[katex]⇒x=30+1 \;\;\;\;\;\;or\;\;\;\;\;\; x=-30+1[/katex]
[katex]so\;\;\;\;\;\; x=31 \;\;\;\;\;\;and \;\;\;\;\;\;x=-29[/katex]
Hence by completing the square
Solution set={31.-29}
you can also see quadratic formula