Question

Please help! If 3x^2-2x+7=0 then (x-1/3)^2

Answer 1

First you need to solve the quadratic and then evaluate the expression
\[\large (x-\frac{1}{3})^{2}\]
Is that the case?

Answer 2

how do I solve it>? @kropot72

Answer 3

You need to use the formula that will solve any quadratic:
\[\large x=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\]
I this case:
a = 3; b = -2 and c = 7.

Answer 4

ok Thanks :)

Answer 5

Note that there is no real solution to the quadratic, the reason being that the discriminant is the square root of -80. Therefore the solution must be imaginary.

Answer 6

what do you mean with imaginary

Answer 7

I got -80?

Answer 8

The square root of a negative number cannot be a real number. Are you sure that you typed the quadratic correctly?
Yes the value of the discriminant is -80. When you take the square root, this is the result:
\[\large \sqrt{-80}=\sqrt{-1}\sqrt{80}=i \sqrt{80}\]
where i represents the square root of negative 1.

Answer 9

yeah

Answer 10

So the solutions to the quadratic are:
\[\large x=\frac{2+8.94i}{6},\ x=\frac{2-8.94i}{6}\]

Answer 11

These solutions can be simplified by dividing each term in the numerator by 6.

Answer 12

If you have not studied imaginary numbers, perhaps this question is not appropriate for you at this time.

Answer 13

Do you have answer choices?

Answer 14

yeah its
A. 20/9
B. 7/9
C. -7/9
D. -8/9
E. -20/9

Answer 15

One of these solutions is correct, which confirms my understanding of the question.

Answer 16

which is?

Answer 17

@undeadknight26

Answer 18

@kropot72 I'm confused on this ._.

Answer 19

To get an exact solution it is best to put the square root of 80 in this form:
\[\large \sqrt{80}=4\sqrt{5}\]
Then the solutions of the quadratic become:
\[\large x=\frac{1}{3}+\frac{2\sqrt{5}i}{3},\ \frac{1}{3}-\frac{2\sqrt{5}i}{3}\]
When each of these two values of x are plugged into the expression
\[\large (x-\frac{1}{3})^{2}\]
we get:
\[\large (\frac{1}{3}+\frac{2\sqrt{5}i}{3}-\frac{1}{3})^{2}=(\frac{2\sqrt{5}i}{3})^{2}=-\frac{20}{9}\]
and
\[\large (\frac{1}{3}-\frac{2\sqrt{5}i}{3}-\frac{1}{3})=(-\frac{2\sqrt{5}i}{3})^{2}=-\frac{20}{9}\]