Question

Solve the quadratic k^2-4k+13=0

Answer 1

Okay, you could use the quadratic formula!
a = 1 (k^2)
b = -4 (-4k)
c = 13

So
\[x = \frac{-b \pm \sqrt{b^2 - 4ac} }{ 2a }\]

Answer 2

use this formula

Answer 3

and put value and get your ans

Answer 4

i got which was wrong

Answer 5

plug in the values for the formula

-b = 4 since -(-4) = 4
b^2 = 16
4(a)(c) = 4 * 1 * 13 = 54
2a = 2
so with that
\[x = \frac{ 4 \pm \sqrt{16 - 54} }{ }\]

Answer 6

5 and -1

Answer 7

over 2

Answer 8

4*1*13 = 52 (correction)

Answer 9

Nono, help her get the answer don't just say it

Answer 10

Oh, oops /)_- thanks mathslover

Answer 11

You're welcome.

Answer 12

k got it so x=-14,22

Answer 13

let me work it out completely

Answer 14

Will you tell me how you got -14 and 22? I got different answers

Answer 15

Okay, so your mistake was at x = -36 / 2
the sqrt sign is looking for a value that multiplied by itself = 36.
so sqrt(36) = 6

Answer 16

... congrats you get \[ \begin{array}l\color{red}{\text{F}}\color{orange}{\text{r}}\color{grey}{\text{e}}\color{green}{\text{e}}\color{blue}{\text{-}}\color{purple}{\text{C}}\color{purple}{\text{u}}\color{red}{\text{p}}\color{orange}{\text{c}}\color{grey}{\text{a}}\color{green}{\text{k}}\color{blue}{\text{e}}\color{purple}{\text{s}}\color{purple}{\text{}}\end{array} \]

Answer 17

oops i forgot it was still square root! thanks