# Completing the square.

• September 23rd 2010, 08:27 AM
greatersanta616
Completing the square.
A level Mathematics. Use the method of completing the square to solve these quadratic equations. Give your answer in the form a(+or-)b root "n" where a and b are rational, and n is an integer.
a. x^2+4x-1=0
d. x^2-8x-3=0

• September 23rd 2010, 08:39 AM
Quote:

Originally Posted by greatersanta616
A level Mathematics. Use the method of completing the square to solve these quadratic equations. Give your answer in the form a(+or-)b root "n" where a and b are rational, and n is an integer.
a. x^2+4x-1=0
d. x^2-8x-3=0

(a)

$(x+p)^2=(x+p)(x+p)=x(x+p)+p(x+p)=x^2+2xp+p^2$

Therefore, in $x^2+4x=1$ the "2p" term is 4, so p is 2.

But... $(x+2)^2=x^2+4x+4$

Hence

$x^2+4x=1\Rightarrow\ x^2+4x+4=1+4$

$(x+2)^2=5$

$x+2=\pm\sqrt{5}$

$x=-2\pm\sqrt{5}$

Try (d)
• September 23rd 2010, 08:46 AM
greatersanta616
(d) x^2-8x-3=0
So..
(x-4)^2-19=0
Therefore.
(x-4)^2=19
henceforth...
x-4=(+or-)root 19
and...
x=4(+or-)root 19

Is this ok?
• September 23rd 2010, 08:49 AM
Quote:

Originally Posted by greatersanta616
(d) x^2-8x-3=0
So..
(x-4)^2-19=0
Therefore.
(x-4)^2=19
henceforth...
x-4=(+or-)root 19
and...
x=4(+or-)root 19

Is this ok?

Yes, that's the completing the square technique..you got it fast!

I think that "a" and "b" being rational, would apply more to the case of using the quadratic formula,
however, those values, while being integers, are rational anyway.
• September 23rd 2010, 08:51 AM
greatersanta616
Quote: