Question

if a=2 ; b=-5; c=5
how would you solve the quadratic equation?

Answer 1

then the two roots of x are:
\(\huge{x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}\)

Answer 2

Use the quadratic foumula

\[x_1 =\frac{ -b + \sqrt{b^2-4ac}} {2a}\]
\[x_2 =\frac{ -b - \sqrt{b^2-4ac}} {2a}\]

Answer 3

ya but i got 4+- the square root of -24 / 4 does this make sense?

Answer 4

your b=-5, how did u get 4+/- ... ?

Answer 5

sry the b=-4 and the c=5

Answer 6

You have imaginary roots. Is that what you're worried about?

Answer 7

yup

Answer 8

Why? Is it a physics problem and you are supposed to have real roots? Because there is nothing wrong with the quadratic formula. Double check your coefficients

Answer 9

u actually have imaginary roots for this..

Answer 10

alright can u tell me where to take it from the imaginary roots?

Answer 11

how did u get square root of -24 ?

Answer 12

because with the coefficients, its (-4)^2-4(2)(5) , in the square root right?

Answer 13

lol! b=-5 not -4

Answer 14

no i i mixed up the coefficents in the original question. i was given a=2;b=-4;c=5

Answer 15

ok,
\(\huge{x=\frac{4 \pm \sqrt{-24}}{4}}=\frac{4 \pm 2\sqrt{-6}}{4}=\frac{2\pm i\sqrt{6}}{2}\)

Answer 16

alright thank you very much!