# check my work on quadratic equations?

• Dec 8th 2009, 06:01 PM
wonderd
check my work on quadratic equations?
can someone check this for me?
• Dec 8th 2009, 06:15 PM
wonderd
• Dec 8th 2009, 06:20 PM
Bacterius
No, both are wrong (first one in the answer, second one in the writing) :(

$\Rightarrow$ In the first one, you must write both answers :

$x = \frac{4}{\sqrt{5}}$ and $x = - \frac{4}{\sqrt{5}}$

$\Rightarrow$ The second one has the correct answer but is not correctly written. Here is an example of how it should be written :

$x^2 - 8x + 5 = 0$

$x^2 - 8x = -5$

$x^2 - 8x + 16 = 11$

$(x - 4)^2 = 11$

$x - 4 = \sqrt{11}$ or $x - 4 = - \sqrt{11}$

$x = \sqrt{11} + 4$ or $x = - \sqrt{11} + 4$

Thus $x = 4 \pm \sqrt{11}$

:)

EDIT : talking about the first question in this post ...
• Dec 8th 2009, 06:24 PM
pickslides
$\frac{-4\pm 2\sqrt{10}}{2}= -2\pm \sqrt{10}$
• Dec 8th 2009, 06:25 PM
jgv115
You must show all your working

What if I didn't know what the quadratic formula was?

You're up to here:

$\frac {-4 \pm 2\sqrt10}{2}$

The next bit is wrong. You didn't factor 2 out of the -4.
• Dec 8th 2009, 06:25 PM
Bacterius
The other question you posted is wrong too :(
I am not taking account of the quadratic equation written on it because I do not see the link with the following.

Here is how you do it :

$\frac{-4 \pm \sqrt{40}}{2} = \frac{-4 \pm \sqrt{4 \times 10}}{2}$

$= \frac{-4 \pm 2 \sqrt{10}}{2} = \frac{2(-2 \pm \sqrt{10})}{2}$ --> This is where you messed up : you did not factorize the $-4$.

$= -2 \pm \sqrt{10}$

Or, more conveniently :

$= 2 \mp \sqrt{10}$
• Dec 8th 2009, 06:37 PM
wonderd
can someone help me understand the factorization part?
• Dec 8th 2009, 06:46 PM
Bacterius
Remember that $ab + ac = a(b + c)$. Here, you have :

$\frac{-4 \pm 2 \sqrt{10}}{2}$

This can be written :

$\frac{-2 \times 2 \pm 2 \sqrt{10}}{2}$

The common factor is two, so you factorize it over the addition. Since you are factorizing all the $2$ (and not $-2$ !), they become $1$ (some people prefer to remove them directly though) :

$\frac{2(-2 \times 1 \pm 1 \sqrt{10})}{2}$

Now you can cancel $2$ out :

$-2 \times 1 \pm 1 \sqrt{10}$

Which is simplified to :

$-2 \pm \sqrt{10}$

Or, even more conveniently :

$2 \mp \sqrt{10}$
• Dec 8th 2009, 07:13 PM
wonderd
still dont really folllow, sry
• Dec 8th 2009, 08:16 PM
jgv115
alright let's break it down.

We have

$-4\pm2\sqrt10$

if we had $2x^2+4x$

we would factorise it by taking 2x out of the equation.

So we divide all terms by 2x:

$x+2$

So same as $-4\pm2\sqrt10$

we can take 2 out of the equation;

So divide everything by 2
you will get: $-2\pm\sqrt10$

I divided the -4 by 2 (which made -2) and 2 by 2 ( which made 1)

Is this more clear now?
• Dec 8th 2009, 08:59 PM
wonderd
i think i got it now, thx a bunch
• Dec 9th 2009, 01:45 AM
jgv115
Cool!
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 36, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}