quadratic equation

**mark** · August 30th 2009, 11:02 AM

hi, i've got the question

by completing the square, solve the following quadratic equation, giving your answer in surd form

$x^2 - x - 1 = 0$

my answer was $x = 0.5 \pm \sqrt (0.75)$

but the books answer is $x = 0.5 \pm \sqrt (5) frac(2)$

could someone please show me a step by step way of coming to this answer
thanks

ps that last bit was supposed to be the square root sign with the 5 to the right of it, then a division bar beneath them and a 2 underneath that, if someone could tell me how to write that properly that would be appreciated too

**e^(i*pi)** · August 30th 2009, 11:25 AM

Originally Posted by mark

hi, i've got the question

by completing the square, solve the following quadratic equation, giving your answer in surd form

$x^2 - x - 1 = 0$

my answer was $x = 0.5 \pm \sqrt (0.75)$

but the books answer is $x = 0.5 \pm \sqrt (5) frac(2)$

could someone please show me a step by step way of coming to this answer
thanks

ps that last bit was supposed to be the square root sign with the 5 to the right of it, then a division bar beneath them and a 2 underneath that, if someone could tell me how to write that properly that would be appreciated too

$x^2 - x -1 =0$

$x^2 - x = 1$

$(x-\frac{1}{2})^2 - \frac{1}{4} = 1$

You should be able to solve it from there. Also recall that $\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$ and also $\sqrt{\frac{a}{c^2}} = \frac{\sqrt{a}}{c}$

**QM deFuturo** · August 30th 2009, 11:59 AM

Originally Posted by mark

but the books answer is $x = 0.5 \pm \sqrt (5) frac(2)$

ps that last bit was supposed to be the square root sign with the 5 to the right of it, then a division bar beneath them and a 2 underneath that, if someone could tell me how to write that properly that would be appreciated too

You were close. You had

[tex]x = 0.5 \pm \sqrt (5) frac(2)[/tex]

and what you want is

[tex]x = 0.5 \pm \frac{\sqrt{5}}{2}[/tex]

which looks like

$x = 0.5 \pm \frac{\sqrt{5}}{2}$

Fractions are written with \frac{N}{D}, where N is the numerator and D is the denominator.

You can always click on any LaTex output in a post and you will get a little popup window that shows the markup used to format it.

**mark** · August 30th 2009, 12:36 PM

i'm still not sure i understand, i'll show my method of figuring it out and hopefully someone will tell me where i went wrong with the equation $x^2 - x - 1 = 0$

i started by putting it like this $(x - 0.5)^2 - 0.75 = 0$ because $-0.5^2$ is -0.25 and then you minus another .75 to get to the original -1 then you turn the equation into $(x - 0.5)^2 = 0.75$ then $(x - 0.5) = \pm\sqrt{0.75}$ then $x = 0.5 \pm \sqrt {0.75}$

**running-gag** · August 30th 2009, 12:56 PM

Hi

i started by putting it like this $(x - 0.5)^2 - 1.25 = 0$ because $(-0.5)^2$ is +0.25 and then you minus another 1.25
to get to the original -1 then you turn the equation into $(x - 0.5)^2 = 1.25$ then $(x - 0.5) = \pm\sqrt{1.25}$ then $x = 0.5 \pm \sqrt {1.25}$
$x = 0.5 \pm \sqrt {1.25} = \frac12 \pm \sqrt {\frac54} = \frac12 \pm \frac{\sqrt {5}}{2}$

**mark** · August 30th 2009, 01:01 PM

ah i see i made a bit of a stupid mistake there, thanks

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