Quadratic and Completing square

**pychon** · October 25th 2010, 03:33 PM

Having some difficulty attempting this problem...

1) $(x-a)^{2} - b^{2} = 0$

Solving:
$(x-a)(x-a) - b^{2} = 0$

$x^{2} - ax - ax - a^{2} - b^{2} = 0$

$x^{2} - 2ax - a^{2} - b^{2} = 0$

... stuck.*

Completing the square

2) $x^{2} + 2bx + c = 0$

.... stuck

**wonderboy1953** · October 25th 2010, 03:44 PM

Originally Posted by pychon

Having some difficulty attempting this problem...

1) $(x-a)^{2} - b^{2} = 0$

Solving:
$(x-a)(x-a) - b^{2} = 0$

$x^{2} - ax - ax - a^{2} - b^{2} = 0$

$x^{2} - 2ax - a^{2} - b^{2} = 0$

... stuck.*

Completing the square

2) $x^{2} + 2bx + c = 0$

.... stuck

For the first problem, try adding $b^2$ to both sides of the equation.

For the second problem, take 1/2 of the coefficient of x, square it and add to both sides of the equation.

Can you go on to solve?

**pychon** · October 25th 2010, 03:46 PM

Ha.. you got to it before I could edit. The first problem is just a quadratic equation, the second is completing the square.

**Archie Meade** · October 25th 2010, 03:51 PM

Originally Posted by pychon

Having some difficulty attempting this problem...

1) $(x-a)^{2} - b^{2} = 0$

Solving:
$(x-a)(x-a) - b^{2} = 0$

$x^{2} - ax - ax - a^{2} - b^{2} = 0$ you should have a +a^2, not a -a^2

$x^{2} - 2ax - a^{2} - b^{2} = 0$

shouldn't you be going

$(x-a)^2=b^2\;\;?$

$x-a= \pm b$

$x=a \pm b$

... stuck.*

Completing the square

2) $x^{2} + 2bx + c = 0$

.... stuck

.

**pychon** · October 25th 2010, 04:04 PM

Originally Posted by Archie Meade

.

Nice catch.... yes, a typo should be -a^2 and yes... should have used the square root instead of factoring the entire thing out (tedious). Correct?

so...

$\sqrt{(x-a)^{2}} = \sqrt{b^{2}}$

$x-a = b$

$x = a \pm b$

Just a quick check if someone wouldn't mind (different problem):

$9x^{2}-6x-2=0$

$9x^{2}-6x=2$

$x^{2}-\frac{2}{3}+\frac{1}{9}=\frac{2}{9}+\frac{1}{9}$

$\sqrt{(x-\frac{1}{3})^{2}} = \sqrt{\frac{3}{9}}$

$x=\frac{1}{3}\pm\frac{\sqrt{5}}{5}$
thanks!

as for completing the square... still trying to burn that into my brain with no luck.

**Archie Meade** · October 25th 2010, 04:24 PM

Originally Posted by pychon

Nice catch.... yes, a typo should be +a^2 and yes... should have used the square root instead of factoring the entire thing out (tedious). Correct?

so...

$\sqrt{(x-a)^{2}} = \sqrt{b^{2}}$

$x-a = b$ take the positive and negative square roots of.. $b^2$ here

$x = a \pm b$

thanks!

as for completing the square... still trying to burn that into my brain with no luck.

The completing the square idea works according to

$(x+q)(x+q)=x(x+q)+q(x+q)=x^2+xq+xq+q^2=x^2+2xq+q^2$

To work backwards from that, you examine the multiplier of x and take half of it...

$x^2+2bx+c=0$

x is multiplied by 2b, so half of that is b, hence we need to introduce $b^2$ by adding to both sides

$x^2+2bx+b^2+c=b^2\Rightarrow\ x^2+2bx+b^2=b^2-c$

Now the left side is a "multiplied out" square

$(x+b)^2=x^2+2bx+b^2$

so you can continue by taking the square root of both sides as before

**Archie Meade** · October 25th 2010, 04:40 PM

Originally Posted by pychon

Just a quick check if someone wouldn't mind (different problem):

$9x^{2}-6x-2=0$

$9x^{2}-6x=2$

$x^{2}-\frac{2}{3}x+\frac{1}{9}=\frac{2}{9}+\frac{1}{9}$

$\sqrt{(x-\frac{1}{3})^{2}} = \sqrt{\frac{3}{9}}$

$x=\frac{1}{3}\pm\frac{\sqrt{5}}{5}$ incorrect, maybe a typo, should have 3 in there

thanks!

$\displaystyle\ x-\frac{1}{3}=\pm\sqrt{\frac{1}{3}}=\pm\frac{1}{\sqr t{3}}\Rightarrow\ x=\frac{1}{3}\ \pm \frac{1}{\sqrt{3}}=\frac{1}{3}\ \pm\frac{\sqrt{3}}{3}=\frac{1\ \pm\sqrt{3}}{3}$

Alternatively...

$9x^2-6x=2\Rightarrow\ (3x)^2-(2)3x+1=2+1$

$(3x-1)(3x-1)=3\Rightarrow\ 3x-1=\pm\sqrt{3}$

$3x=1 \pm \sqrt{3}\Rightarrow\ x=\displaystyle\frac{1\pm\ \sqrt{3}}{3}$

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