Question

Find the solutions of the equation.

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

Answer 1

Is it 1/2x^2 or 1/ (2x^2-x+5)

Answer 2

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

Answer 3

Is the 1/ part of 2x^2 or of the whole expression?

Answer 4

Part of

Answer 5

1/2x^2 - x + 5 = 0
1/2x^2 = x - 5
1 = 2x^3 - 10x^2
2x^3 - 10x^2 - 1 = 0

Solve using a cubic equation solver

Answer 6

multiply every term by 2 gives

\[x^2 - 2x + 10 = 0\]

the quadratic can't be factored so use the general quadratic formula

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

now substitute a = 1, b = -2 and c = 10 and evaluate

Answer 7

Working on it.....

Answer 8

Yeah I cant finish the evaluation

Answer 9

ok so you have

\[x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4\times 1 \times 10}}{2 \times 1} = \frac{-2 \pm \sqrt{4 - 40}}{2}\]

which becomes

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

so - 36 can be written as

\[36i^2\]

so substituting you get

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

now, whoever wrote the answers to this question has done an extremely poor job.

the correct answer is

\[x = -1 \pm 3i\]

to get one of your answer choices they have rewritten it as

\[\frac{-2 \pm \sqrt{4}\times \sqrt{9} \times \sqrt{i^2}}{2} = \frac{-2 \pm2\sqrt{9}i}{2}\]

cancel the common factor of 2 for your answer...

my problem is why leave \[\sqrt{9}\] why not write 3.

anyway, hope this helps

Answer 10

That was incredibly helpful. Thanks alot

Answer 11

oops should be

\[\frac{2 \pm \sqrt{9}i}{2}\]

I missed a negative... just saw it when checking