Another square technique...

Printable View

May 20th 2010, 07:01 AM
tysonrss

Another square technique...

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.
May 20th 2010, 07:16 AM
harish21

Quote:

Originally Posted by tysonrss

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.

$2x^{2} + 32x + 12 = 0$

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

$x^{2} + 16x + 6 = 0$

$x^2 + 2\times 8\times x + 6 = 0$

Can you take it further? Please try
May 20th 2010, 07:26 AM
tysonrss

Quote:

Originally Posted by harish21

$2x^{2} + 32x + 12 = 0$

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

$x^{2} + 16x + 6 = 0$

$x^2 + 2\times 8\times x + 6 = 0$

Can you take it further? Please try

Would it be (x + 6)2 = 58 ?
May 20th 2010, 07:43 AM
harish21

Quote:

Originally Posted by tysonrss

Would it be (x + 6)2 = 58 ?

No.

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

So $(x+c)^2 = x^2+ 2.x.c +c^2$ ...(II)

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

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

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

To make this $x^2+ 2\times x \times 8 + 64 = 0$ ,

you need to add 58 right?

so,

$x^2+ 2\times x \times 8 + 6 +58 -58= 0$

Now put the above equation in the form of $(x+c)^2=a$
May 20th 2010, 07:55 AM
tysonrss

Quote:

Originally Posted by harish21

No.

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

So $(x+c)^2 = x^2+ 2.x.c +c^2$ ...(II)

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

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

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

To make this $x^2+ 2\times x \times 8 + 64 = 0$ ,

you need to add 58 right?

so,

$x^2+ 2\times x \times 8 + 6 +58 -58= 0$

Now put the above equation in the form of $(x+c)^2=a$

$(<i>x</i> + 8)2 = 58$ would be it than right?
May 20th 2010, 08:08 AM
harish21

Quote:

Originally Posted by tysonrss

$(<i>x</i> + 8)2 = 58$ would be it than right?

yes