Question on Completing the X

**Lites** · August 16th, 2010, 11:20 PM

Hello there, I have a question on how to complete the x on a quadratic equation which has the coefficient of a. It cannot be solved using factorization. The equation is:

$8x^2 + 4x - 1 = 0$

Thank You

**Prove It** · August 16th, 2010, 11:45 PM

I assume you mean to complete the square...

$8x^2 + 4x - 1 = 0$

$8\left(x^2 + \frac{1}{2}x - \frac{1}{8}\right) = 0$

$8\left[x^2 + \frac{1}{2}x + \left(\frac{1}{4}\right)^2 - \left(\frac{1}{4}\right)^2 - \frac{1}{8}\right] = 0$

$8\left[\left(x + \frac{1}{4}\right)^2 - \frac{3}{16}\right] = 0$

$8\left(x + \frac{1}{4}\right)^2 - \frac{3}{2} = 0$ .

Go from here.

**Lites** · August 17th, 2010, 02:53 AM

Originally Posted by Prove It

I assume you mean to complete the square...

$8x^2 + 4x - 1 = 0$

$8\left(x^2 + \frac{1}{2}x - \frac{1}{8}\right) = 0$

$8\left[x^2 + \frac{1}{2}x + \left(\frac{1}{4}\right)^2 - \left(\frac{1}{4}\right)^2 - \frac{1}{8}\right] = 0$

$8\left(x + \frac{1}{4}\right)^2 - \frac{3}{16}\right] = 0$

$8\left(x + \frac{1}{4}\right)^2 - \frac{3}{2} = 0$ .

Go from here.

Thank you but I'm little bit confused on how do you exactly have

$- \frac{3}{2}$

From

$\frac{3}{16}$

I've also noticed that the second from the last line did not show up, so I edit your formatting a little bit. I'm not sure if what I'm doing is right. Can you verify?

**Prove It** · August 17th, 2010, 02:57 AM

I expanded the outermost (square) brackets.

$8\left(-\frac{3}{16}\right) = -\frac{3}{2}$ .

**Lites** · August 17th, 2010, 03:08 AM

Originally Posted by Prove It

I expanded the outermost (square) brackets.

$8\left(-\frac{3}{16}\right) = -\frac{3}{2}$ .

Ah I see, can you give me a bit of insight how to find x? Or at least to eliminate the 8?

**Prove It** · August 17th, 2010, 03:12 AM

Use the standard method of "undoing" everything using inverse operations.

**Lites** · August 17th, 2010, 03:48 AM

Is it possible to do this:

$8(x+1/4)^2 = 3/2$

$8(x+1/4) = root(3/2)$

$8x + 2 = root(3/2)$

$8x = root(3/2) - 2$

$x = (root(3/2) - 2) / 8$

But it gives up quite weird answer...

I'm sorry also because of bad formattig I do not know how to operate Latex properly.

**Prove It** · August 17th, 2010, 03:54 AM

No, when you undo everything you go in the reverse of the order of operations.

Since Exponentiation comes before Multiplication, when you go in reverse, the Multiplication is undone before the Exponentiation...

**Lites** · August 17th, 2010, 04:09 AM

Originally Posted by Prove It

No, when you undo everything you go in the reverse of the order of operations.

Since Exponentiation comes before Multiplication, when you go in reverse, the Multiplication is undone before the Exponentiation...

Allright, would this be possible:

$8(x+1/4)^2 = 3/2$

$(x+1/4)^2 = 8(3/2)$

$(x+1/4) = root(12)$

$x = root(12)(1/4)$

**Prove It** · August 17th, 2010, 04:22 AM

How do you undo multiplication?

**Lites** · August 17th, 2010, 04:28 AM

Originally Posted by Prove It

How do you undo multiplication?

By dividing it I suppose, I'm really sorry if action looks like an imbecible. I just want to understand it. Just having problem with that number 8.

**Prove It** · August 17th, 2010, 04:29 AM

Yes, you need to divide both sides by 8.

What is $\frac{3}{2} \div 8$ ?

**Lites** · August 17th, 2010, 04:46 AM

Originally Posted by Prove It

Yes, you need to divide both sides by 8.

What is $\frac{3}{2} \div 8$ ?

AH I see would it be like this then?

$(x+\frac{1}{4})^2 = \frac{3}{16}$
$(x+\frac{1}{4}) = \sqrt{\frac{3}{16}}$
$(x+\frac{1}{4}) = \frac{\sqrt{3}}{4}$
$x = \frac{1}{4}(1-\sqrt{3})$

**Prove It** · August 17th, 2010, 05:08 AM

Originally Posted by Lites

AH I see would it be like this then?

$(x+\frac{1}{4})^2 = \frac{3}{16}$
$(x+\frac{1}{4}) = \sqrt{\frac{3}{16}}$
$(x+\frac{1}{4}) = \frac{\sqrt{3}}{4}$
$x = \frac{1}{4}(1-\sqrt{3})$

Very close. There are two square roots to every nonnegative number, one positive and one negative.

So that means it should read...

$\left(x + \frac{1}{4}\right)^2 = \frac{3}{16}$

$x + \frac{1}{4} = \pm \sqrt{\frac{3}{16}}$

$x + \frac{1}{4} = \pm \frac{\sqrt{3}}{4}$

$x = -\frac{1}{4} \pm \frac{\sqrt{3}}{4}$ .

So your two solutions are $x = \frac{-1 - \sqrt{3}}{4}$ and $x = \frac{-1 + \sqrt{3}}{4}$ .

**Lites** · August 17th, 2010, 07:28 AM

Originally Posted by Prove It

Very close. There are two square roots to every nonnegative number, one positive and one negative.

So that means it should read...

$\left(x + \frac{1}{4}\right)^2 = \frac{3}{16}$

$x + \frac{1}{4} = \pm \sqrt{\frac{3}{16}}$

$x + \frac{1}{4} = \pm \frac{\sqrt{3}}{4}$

$x = -\frac{1}{4} \pm \frac{\sqrt{3}}{4}$ .

So your two solutions are $x = \frac{-1 - \sqrt{3}}{4}$ and $x = \frac{-1 + \sqrt{3}}{4}$ .

Finally, I got it!

Thank you for your kind help Prove it!

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