# Math Help - Quadratic equations using the formula method

1. ## Quadratic equations using the formula method

Can anyone solve the following quadratic equation using the formula method...

Xsquared - 10x + 3 = 0

Please show working out so it is possible for me to follow and see where i am going wrong

Thanks

2. Originally Posted by duckegg911
Can anyone solve the following quadratic equation using the formula method...

Xsquared - 10x + 3 = 0

Please show working out so it is possible for me to follow and see where i am going wrong

Thanks
Let me try again...

$x^2 - 10x + 3 = 0$

It seems he's allowed to use the quadratic formula.

All quadratic equations are written in the format of $ax^2 + bx + c = 0$

That means that $a$ is the co-efficient of the $x^2$ ; $b$ is the co-efficient of $x$ ; and $c$ is the constant.

$x = \frac{ -b \pm \sqrt{b^2 - 4ac} }{2a}$

All you have to do is substitute the corresponding co-efficients into the formula.

3. A standard quadratic equation is in the form
$ax^2 + bx + c = 0$
a, b and c being the coefficients.

The formula for the discriminant ( $\Delta$) is
$\Delta = b^2 - 4ac$

If $\Delta > 0$, there are 2 different roots.
If $\Delta = 0$, there are 2 equivalent roots. ( $x_1=x_2$)
If $\Delta < 0$, there are no real roots.

The roots are,

$x_1 = \frac{-b - \sqrt{\Delta}}{2a}$

$x_2 = \frac{-b + \sqrt{\Delta}}{2a}$

Now plug a, b and c in these formulas.

4. Let's say we write your equation this way.

$ax^2+bx+c=0$
where:
$a=1$
$b=-10$
$c=3$

now we have the coefficients we use in the equation.

$x_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Now you just put a,b and c into the equation and that's it.

There are usually 2 results for x.

5. Sorry, was writing the answer while you posted them.

6. Originally Posted by Pinsky
Sorry, was writing the answer while you posted them.
No problem, at least now (s)he has 3 explanations to choose from...

7. Originally Posted by Pinsky
Sorry, was writing the answer while you posted them.
Thanks to you both, ill take a long hard look at these explanations and let you both know how i get on ,

thanks again

8. Originally Posted by janvdl
No problem, at least now (s)he has 3 explanations to choose from...
Not to mention the explanations accompanying the examples s/he should have in their class notes .....

9. Originally Posted by janvdl
Let me try again...

$x^2 - 10x + 3 = 0$

It seems he's allowed to use the quadratic formula.

All quadratic equations are written in the format of $ax^2 + bx + c = 0$

That means that $a$ is the co-efficient of the $x^2$ ; $b$ is the co-efficient of $x$ ; and $c$ is the constant.

$x = \frac{ -b \pm \sqrt{b^2 - 4ac} }{2a}$

All you have to do is substitute the corresponding co-efficients into the formula.
Are the answers for -9.6904 and -3.095 ???

10. Originally Posted by duckegg911
Are the answers for -9.6904 and -3.095 ???

11. Originally Posted by janvdl
This is because - plus a - = a + ???

12. Originally Posted by duckegg911
This is because - plus a - = a + ???
sorry that sound silly....

the answers you have are 9.6904 and 3.095 ?

13. After substitution into the formula, your formula should look like this:

$x = \frac{ +10 \pm \sqrt{100 - 4(1)(3)} }{2}$

Does it?

14. Here's the full solution. Ask me about any step you do not understand/

$x = \frac{ +10 \pm \sqrt{100 - 4(1)(3)} }{2}$

$x = \frac{ +10 \pm \sqrt{100 - 12} }{2}$

$x = \frac{ +10 \pm \sqrt{88} }{2}$

BUT $\sqrt{88} = \sqrt{4} \times \sqrt{22}$

AND we know $\sqrt{4} = 2$

$x = \frac{ +10 \pm \sqrt{4} \cdot \sqrt{22} }{2}$

$x = \frac{ +10 \pm 2 \sqrt{22} }{2}$

Divide 10 by 2, and divide $2 \sqrt{22}$ by 2

$x = +5 \pm \sqrt{22}$

15. Originally Posted by janvdl
Here's the full solution. Ask me about any step you do not understand/

$x = \frac{ +10 \pm \sqrt{100 - 4(1)(3)} }{2}$

$x = \frac{ +10 \pm \sqrt{100 - 12} }{2}$

$x = \frac{ +10 \pm \sqrt{88} }{2}$

BUT $\sqrt{88} = \sqrt{4} \times \sqrt{22}$

AND we know $\sqrt{4} = 2$

$x = \frac{ +10 \pm \sqrt{4} \cdot \sqrt{22} }{2}$

$x = \frac{ +10 \pm 2 \sqrt{22} }{2}$

Divide 10 by 2, and divide $2 \sqrt{22}$ by 2

$x = +5 \pm \sqrt{22}$
I have -10 all the way through rather than +10 ???

Page 1 of 2 12 Last
{"extraction_info": {"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": 49, "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, "math_score": 0.8962874412536621, "perplexity": 912.8069303589814}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-22/segments/1432207928907.65/warc/CC-MAIN-20150521113208-00330-ip-10-180-206-219.ec2.internal.warc.gz"}