Question

Need some help
25w^2 -25w=-3

Answer 1

\[25w^2-25w=-3\]
\[w^2-w =-\frac{3}{25}\]
\[w^2-w+\frac{1}{4}=\frac{13}{100}\]
\[(w-\frac12)^2=\frac{13}{100}\]
sqrt and solve for w, you should find:
\[w=\frac12\pm\frac{\sqrt{13}}{10}\]

Answer 2

Ok I solved using quadratic equation.. I got the answer \[w=\frac{ 1\pm \sqrt{13} }{ 5 } \]

Did I do something wrong?

Answer 3

I don't use the quadratic, however our answer are very similar--but i would make sure you did your division and factoring accurately as it seems like we are a bit off from one another

Answer 4

Alright I will work on it again.

Answer 5

solved for w

Answer 6

I have check my work again and got the final answer (I made 1 mistake)\[w=\frac{ 1\pm \sqrt{13}}{ ? }\]

Answer 7

over 2 i mean not ?

Answer 8

@SpecialMathStudent Couldn'y you simplify the 5/10 by making 1/2? Because that is what I did.

Answer 9

Yeah most likely, if its asking you to reduce it to the lowest form.

Answer 10

\[w=\frac{25\pm \sqrt{625-300}}{50}=\frac{25\pm \sqrt{325}}{50}=\frac{25\pm5\sqrt{13}}{50}=\]

Answer 11

\[\frac{5\pm \sqrt{13}}{10}\]

Answer 12

@Mertsj So reducing that to \[\frac{1\pm \sqrt{13}}{2}\] is the wrong answer?

Answer 13

Or you could split it up into 2 fractions and write:
\[\frac{1}{2}\pm\frac{\sqrt{13}}{10}\]

Answer 14

Id say most likely, stick with the 5 and 10 that should be your answer.

Answer 15

Yes. That is wrong.

Answer 16

Thank you!

Answer 17

Yeap!

Answer 18

yw