Help! I think I have this figured out but I wanna know for sure. Solve using the Quadratic Form. 3x^2+5x=2
What was your solution?
I got 4,6
That's not what I'm getting
Did you first subtract the 2?
Yes, that's what I was unsure about. I didn't know if that was the correct way to do it. Or if you make it a perfect square.
Using the quadratic formula you only have to plug in values for a, b, and c, and simplify. So you should be simplifying this expression\[x={{{-5}\pm \sqrt{5^2-4(3)(-2)}}\over2(3)}\]
I can check your answers again for you if that would help.
So basically, you just input whats in the main equation?
And yes please (:
Exactly. The coefficient of your \(x^2\) term is \(a=3\), the coefficient of your \(x\) term is \(b=5\), and the coefficient of your constant term is \(c=-2\). Then you just use those to plug it into the quadratic equation.
Do you have to divide by 2 (a) in this case it would be 2(3)=6 so the top divided by six.
Yes.
I got 2/3 & 1...
Close, but not quite.
What'd I do wrong? :(
I'm not sure. Let's do this one step at a time. What's\[5^2-4(3)(-2)?\]
25+24?
Looks correct.
Thats where I got mixed up. I multiplied wrong. i had 25-24 and got one instead of 49..
That would have screwed things up.
Yeah, it did! Well I am pretty positive I got it this time. -1/3 , 2
Switch the negatives, and it'll look perfect.
Remember that it's a \(-5\) not a \(+5\) on top.
Okay, Thank you. Will you check the next problem i did?
Sure.
x^2+16=8x And i got just 4
4 looks correct to me also. Good job!
How about x(2x-5)=12 I got 5\[\pm \sqrt{-71}\div4\]
theres suppose to be a 5 infront
That's not what I'm getting. Did you remember to distribute the x , and then subtract the 12, then use the quad. formula?
-3/2, 4 ?
I keep making simple mistakes,
That looks right to me. As for the simple mistakes, it's better to make them now rather than on a test. I still make many simple mistakes even though I've done all these things before.
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