Question

Find the solutions of the equation.

1/2x2 - x + 5 = 0

Answer 1

A. \[1 \pm \sqrt{9}i \]
B. \[1 \pm \sqrt{11}i\]
C. \[-1 \pm \sqrt{9}i\]
D. \[-1 \pm \sqrt{11}i\]

Answer 2

it can be rewritten as::
\[1-2x ^{3}+10x ^{2}=0\]
or
\[2x ^{3}-10x ^{2}-1=0\]

Answer 3

now could you show me the steps to solve that?

Answer 4

find out the first root by hit and trial method

Answer 5

im not really sure how to do that

Answer 6

1/2x^2 - x + 5 = 0


x^2 - 2x + 10 = 0 ... multiply every term by 2


Now use the quadratic formula to solve for x


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


\[\Large x = \frac{-(-2)\pm\sqrt{(-2)^2-4(1)(10)}}{2(1)}\]


I'll let you finish. Tell me what you get

Answer 7

i got b some how.

Answer 8

b is incorrect, try it again and post your work if you can

Answer 9

im not sure how to really solve it but im trying.
i simplified the numbers under the radicand and got 33?
then did 2(1) = 2
IM SO CONFUSED -_-

Answer 10

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


\[\Large x = \frac{-(-2)\pm\sqrt{(-2)^2-4(1)(10)}}{2(1)}\]


\[\Large x = \frac{2\pm\sqrt{4-(40)}}{2}\]


\[\Large x = \frac{2\pm\sqrt{-36}}{2}\]


\[\Large x = \frac{2+\sqrt{-36}}{2} \ \text{or} \ x = \frac{2-\sqrt{-36}}{2}\]


\[\Large x = \frac{2+6*i}{2} \ \text{or} \ x = \frac{2-6*i}{2}\]


\[\Large x = 1+3*i \ \text{or} \ x = 1 - 3i\]

Answer 11

now because sqrt(9) = 3

we can write the answer as \[\Large x = 1\pm \sqrt{9}*i\]

Answer 12

Omg Jim thank you so much.. I understand now (:

Answer 13

you're welcome