Solve the following system of equations and show all work. y = 2x^2 y = –3x −1
To start you have to put them equal to each other then you have to get everything to the one side so one side has everything and the other side is equal to zero
so 2x^2=-3x-1 ? @taramgrant0543664
yes that is the first step! so now the next step is to get it all on to one side (I suggest that you move the stuff on the right to the left side so you don't have to deal with the negatives)
2x^2+3x+1=0 ? @taramgrant0543664
Perfect! So now the next step that we are going to do is put it into the quadratic formula: a will represent the the number in front of the x^2 in this case 2 b will represent the number in front the x in this case 3 and c will represent the last number in this case 1\[(-b \pm \sqrt{b ^{2}-4ac})\div2a\]
You will get two numbers one when using the + and one when using the - from the +/-
-1/2 and -1 ?? @taramgrant0543664
Yep that's what I got!
cool thanks , can you help with more ?
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