Another square technique...

May 2010
43
0
Use the technique of completing the square to transform the quadratic equation into the form (x + c)2 = a.

2x2 + 32x + 12 = 0

Thanks alot guys.
 
Feb 2010
1,036
386
Dirty South
Use the technique of completing the square to transform the quadratic equation into the form (x + c)2 = a.

2x2 + 32x + 12 = 0

Thanks alot guys.
\(\displaystyle 2x^{2} + 32x + 12 = 0\)

\(\displaystyle 2(x^{2} + 16x + 6 = 0)\)

\(\displaystyle x^{2} + 16x + 6 = 0\)

\(\displaystyle x^2 + 2\times 8\times x + 6 = 0\)

Can you take it further? Please try
 
May 2010
43
0
\(\displaystyle 2x^{2} + 32x + 12 = 0\)

\(\displaystyle 2(x^{2} + 16x + 6 = 0)\)

\(\displaystyle x^{2} + 16x + 6 = 0\)

\(\displaystyle x^2 + 2\times 8\times x + 6 = 0\)

Can you take it further? Please try
Would it be (x + 6)2 = 58 ?
 
Feb 2010
1,036
386
Dirty South
Would it be (x + 6)2 = 58 ?
No.

You are left with \(\displaystyle x^2+ 2\times x \times 8 + 6 = 0\).....(I)

So \(\displaystyle (x+c)^2 = x^2+ 2.x.c +c^2\)...(II)

if you compare I and II, you should see that c = 8.

So you need to have \(\displaystyle (x+8)^2\), which is equal to \(\displaystyle x^2 + 2.x.8+ 64\)

now if you look at equation (I) again, you already have \(\displaystyle x^2+ 2\times x \times 8 + 6 = 0\).

To make this \(\displaystyle x^2+ 2\times x \times 8 + 64 = 0\),

you need to add 58 right?

so,

\(\displaystyle x^2+ 2\times x \times 8 + 6 +58 -58= 0\)

Now put the above equation in the form of \(\displaystyle (x+c)^2=a\)
 
May 2010
43
0
No.

You are left with \(\displaystyle x^2+ 2\times x \times 8 + 6 = 0\).....(I)

So \(\displaystyle (x+c)^2 = x^2+ 2.x.c +c^2\)...(II)

if you compare I and II, you should see that c = 8.

So you need to have \(\displaystyle (x+8)^2\), which is equal to \(\displaystyle x^2 + 2.x.8+ 64\)

now if you look at equation (I) again, you already have \(\displaystyle x^2+ 2\times x \times 8 + 6 = 0\).

To make this \(\displaystyle x^2+ 2\times x \times 8 + 64 = 0\),

you need to add 58 right?

so,

\(\displaystyle x^2+ 2\times x \times 8 + 6 +58 -58= 0\)

Now put the above equation in the form of \(\displaystyle (x+c)^2=a\)
\(\displaystyle (x + 8)2 = 58\) would be it than right?