Thread: quadratic equation

hi, i've got the question

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

$\displaystyle x^2 - x - 1 = 0$

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

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

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

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

$\displaystyle x^2 - x - 1 = 0$

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

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

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

$\displaystyle x^2 - x -1 =0$

$\displaystyle x^2 - x = 1$

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

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

Originally Posted by mark

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


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

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

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.

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 $\displaystyle x^2 - x - 1 = 0$

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

Hi

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

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