Question

Solve x2 – 7x = –13

Answer 1

Plug it into the quadratic formula.
\[-b ± \sqrt{b^{2} - 4ac}\over2a\]
where a = 1
b = -7
c = 13

Answer 2

you can solve this by using the quadratic formula
first I would put it in a,b,c,d format
then i would have x^2-7x+13=0

a= 1
b=-7
c=13

b^2-4ac
(-7)^2-4(1)(13)= -3

which gives you a Real Irrational # because its a real number less than 0, and its not a perfect square

Answer 3

is it :

Answer 4

use completing the square

\[x^2 - 7x + (\frac{7}{2})^2 = -13 + (\frac{7}{2})^2\]
\[x^2 - 7x + \frac{49}{4} = -13 + \frac{49}{4}\]
\[(x - \frac{7}{2})^2 = \frac{-3}{4}\]
\[x - \frac{7}{2} = \pm \sqrt{\frac 34}\]
\[x = \frac72 \pm \sqrt{\frac 34}\]

Answer 5

i mean \[x = \frac 72 \pm \sqrt{-\frac 34}\] i forgot the negative sign

Answer 6

is that even possible?? wouldnt it be kinda like wht i drew??

Answer 7

well yours is right actually..you see if i simplify this \[x = \frac 72 \pm \frac{\sqrt{-3}}{2}\]
\[x = \frac{ 7 \pm \sqrt{-3}}{2}\]
\[x = \frac{7 \pm i \sqrt 3}{2}\]

i just thought i shouldn't give the answer i didnt know you posted yours

Answer 8

haha ok yah!! i thought i did it wrong! :P

Answer 9

well...7 should be positive

Answer 10

remember in quadratic formula..it's -b

therefore -(-7) = +7

Answer 11

oh ok i allways forget about my signs!! thanks!