Question

Use the quadratic formula to find the solution to the quadratic equation given below
3x^2-5x+1=0

Answer 1

can someone help me on my question please

Answer 2

\[x=\frac{5\pm \sqrt{37} }{ 6 }\]

Answer 3

is that the answer? thats what i got

Answer 4

that's what I got too

Answer 5

I think you have a sign error under the radical.

Answer 6

CAN SOMEONE PLEASE HELP ME!!!!!!

Answer 7

its suppose to be like that -_-

Answer 8

is it right mathstudent 55

Answer 9

I say it is [5 +- sqroot(13)] / 6

Answer 10

\(3x^2-5x+1=0\)

\(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

\(x = \dfrac{-(-5) \pm \sqrt{(-5)^2 - 4(3)(1)}}{2(3)} \)

\(x = \dfrac{5 \pm \sqrt{25 - 12}}{6} \)

\(x = \dfrac{5 \pm \sqrt{13}}{6} \)

Answer 11

@wolf1728 is correct.

Answer 12

The discriminant is 25 - 12 = 13, not 25 + 12 = 37.

Answer 13

oh

Answer 14

By completing the square......\[3x^2-5x+1=0 \implies x^2-\frac{5}{3}x=\frac{-1}{3}\]\[\implies x^2-\frac{5}{3}x+\frac{25}{36}=\frac{25}{36}-\frac{1}{3} \implies (x-\frac{5}{6})^2=\frac{13}{36}\]\[\implies x-\frac{5}{6}= \pm \frac{\sqrt{13}}{6} \implies x=\frac{-5 \pm \sqrt{13}}{6}\]

Answer 15

so my answers are \[x=\frac{ 5+\sqrt{13} }{ 6 }x=\frac{ 5-\sqrt{13} }{ 6 }\]

Answer 16

guys im conffused with the answers

Answer 17

@romanortiz65, your final answers are correct.

@animalain, you made a mistake in your final result.

Answer 18

I hate sign errors. Thanks for pointing it out.