OpenStudy (anonymous):

i just need the answer choice. I know how to get it. just not sure. Solve. 3x2 = -15x + 21 answer choices: the quantity of negative 5 plus or minus i square root 3 all over 2 the quantity of 5 plus or minus square root 53 all over 2 the quantity of 5 plus or minus 3 square root 3 all over 2 the quantity of negative 5 plus or minus square root 53 all over 2

OpenStudy (mertsj):

1. Get it equal to 0 2. Divide both sides by 3 3. Identify a, b, and c 4. Plug into the quadratic ormula.

OpenStudy (anonymous):

which one is it though...?

OpenStudy (whpalmer4):

Follow those steps. Do you need help doing so?

OpenStudy (whpalmer4):

Those steps are exactly how we would find the answer to give to you, so why don't you do it and get the practice?

Show your work for step 4, then we can analyze your problem.

OpenStudy (anonymous):

okay, question, do I put the right side equal to zero or he left?

OpenStudy (mertsj):

The right.

OpenStudy (anonymous):

okaaaay..x=7/5?

OpenStudy (mertsj):

Could you show how you got that?

OpenStudy (anonymous):

-15x+21=0 15x=-21 x=7/5

OpenStudy (whpalmer4):

You misunderstood the directions: you need to rearrange the equation so that everything is the left, and 0 on the right. You still have all three terms!

OpenStudy (anonymous):

OHHH! thanks!

OpenStudy (whpalmer4):

:-)

OpenStudy (anonymous):

so its going to be 3x^2+15x-21=0

OpenStudy (whpalmer4):

yes

OpenStudy (anonymous):

then what do I do?

OpenStudy (mertsj):

Step 2

OpenStudy (whpalmer4):

Now you have your quadratic in the form desired: ax^2+bx+c = 0. plug values into quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

OpenStudy (mertsj):

Make your life simpler and divide both sides by 3 first.

OpenStudy (anonymous):

do I not simplify them because they are divisible by 3..?

OpenStudy (whpalmer4):

you can if you want. makes no difference to the answer.

OpenStudy (anonymous):

so I got s(x^2+5x-7)=0

OpenStudy (anonymous):

supposed to be a 3 in the begining

OpenStudy (whpalmer4):

yeah. so now your values of a,b,c are?

OpenStudy (anonymous):

a=3 b=5 c=-7?

OpenStudy (anonymous):

a=1

OpenStudy (whpalmer4):

you divide by 3 to get rid of that leading 3... right, a = 1, b = 5, c = -7

OpenStudy (anonymous):

thanks so much! I got it!

OpenStudy (whpalmer4):

which choice is the correct answer?

OpenStudy (anonymous):

D

OpenStudy (whpalmer4):

very good!

OpenStudy (anonymous):

:)!!!!!!!!

OpenStudy (whpalmer4):

go forth and solve many quadratics :-)

OpenStudy (whpalmer4):

enjoy the fact that you get that nice, simple formula to use. It all goes out the window when you start doing x^3...