Question

Help with Pre-Cal?

Answer 1

posting specifics helps

Answer 2

It says to solve algebraically and confirm graphically, and the first problem is \(v^2-5=8-2v^2\)

Answer 3

I was getting there, lol XD

Answer 4

\(\bf v^2-5=8-2v^2\implies v^2+2v^2=8+5\implies v=?\)

Answer 5

put the variables on one side constants on the other
keep the balance of the equation

Answer 6

When I solved aglebraically I got \(v=\sqrt{\frac{13}{3}}\)

Answer 7

But when I graphed it, I don't think it's right... but I may have made an error. That's why I'm asking for help.

Answer 8

oops, I meant to say when I graphed it, it didn't match my algebraic solution.

Answer 9

notice, is a quadratic equation, 2nd degree polynomial, thus 2 answers
\(\bf v^2-5=8-2v^2\implies v^2+2v^2=8+5\implies v=\pm\sqrt{\cfrac{13}{3}}\)

Answer 10

I forgot to put that, but I know that. lol

Answer 11

hmmm ok.... so.. whathmm how did you do the graph anyway? just a table of values?

Answer 12

I set the equations equal to zero

Answer 13

and used a calculator

Answer 14

equation**

Answer 15

so.... I'd think... the graph is ok...no?

Answer 16

obviously, I got a parabola with the y-intercept -13. And my zeroes were \(\pm\)2.082

Answer 17

Is that correct?

Answer 18

\(\bf \pm\sqrt{\cfrac{13}{3}}=\pm2.08166599946613273528\) so, the zeros look fine

Answer 19

ohhhh... so I was right, sorry. I was thinking of something else when I graphed it and that was my problem lol

Answer 20

Thanks *facepalm*