solve for r r^2+3r=8 Example r^2+3r=1 r=-3+√13/2,-3-√13/2
looks like the quadratic equation might help here
r^2 +3r -8 =0 a=1 b=3 c=-8 Use the quadratic formula to get your answer(s) for x.
The quadratic equation is where my problem lies
Do you know what the quadratic formula is?
((-1) +/- sqrt ((3)^2-(4)(1)(-8)))/((2)(1))
(-1 +/- sqrt (41))/2 ***THAT MEANS*** x = (-1 + sqrt(41))/2 x = (-1 - sqrt(41))/2
Thanks so much
This is what you do whenever factoring is impossible.
Or if you're too lazy to factor :-)... like I usually am LOL
so how would i put x^2+7x=10 into the quadratic equation
Okay so your format is going to be ax^2+bx+c So with x^2+7x=10, you always always always get everything on the left side. You're setting it equal to zero. x^2+7x-10=0 Now what are a, b, and c?
sorry typed problem wrong x^2+7x-10=0
k... so what do you think a, b, and c are?
a is 1 b is 7 c is 10
Yup!
I'll ask you one now. -x^2+2x+3 What is "a" in this case?
-1
yep
my problem is keeping it all in order
Now if there's a term, say b=-2 and you need to plug it into your formula, the first thing you're putting in is -b. So how to do that: -(-2). So that will be a +2.
I understand. And I would recommend to do the problem in as many steps as you need to. And to slow down and pay close attention. I took an engineering problems course last quarter and I made a lot of mistakes when I didn't pay attention :-)
You can try that problem and we'll compare answers if you want.
ok is it -7+√89/2,-7-√89/2
That's what I got too. You know what you're doing!
You have no idea how much you have helped me thanks again!!!!!
I'm glad :-) So step 1: Get everything on the left side so that it's set equal to zero. step 2: write down what your a, b, and c values are step 3: plug in to the formula and go slow
I truly cannot thank you enough I have been struggling for days
I would recommend Paul's online notes as an alternative to your book. Most algebra books truly suck. http://tutorial.math.lamar.edu/Classes/Alg/SolveQuadraticEqnsII.aspx
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