using the quadratic formula solve each equation. give two forms of each solution an expressionwith radical and a calculator approximation rounded off to two decimal places. 2x^2+3x-4=0.
What part of this problem do you need help with?
the first part
Ive not done these before
do you know the quadratic equation?
Yes
so for any 2nd degree polynomial ax^2 + bx +c = 0 In your case a=2, b=3, and c = -4
(-b +- sqrt(b^2-4ac))/2a
its basically plug and chug
try putting the values of a, b, and c into the quad equation and tell me what you get or at least let me see the substitution and then Ill help you simplify if needed
-3+-Sqrt3^2-3(2)(-4)
this is as far as i have gotten
/3 (?)
cool deal! So lets go ahead and make the 3^2 into 9, and that -3(2)(-4) into +24 So youve got -3 +- sqrt(23) all over 4. I think that's about as far as you can take it, considering that 23 is prime
Do you know how to put that into your calc to get an estimate?
I don't know I will try
how do you do the plus or minus on it ?
You just have to do it twice, once with plus and once with minues
oh ok
ok I got 2.74456
for the + or the -?
so basically, when you are solving a quadratic equation, you are finding the x values for which y = 0
that is +
ok then cool so you understand that quadratic equations have 2 solutions?
do I do the sqrt of 23 or the whole equation?
Then with the - I got -17.2336...
the sqrt of 23 only
oh did not know that thanks
it should be (3/4) - sqrt(23)/4 dont forget the /2a on the bottom
right
ok this is what i got -3sqrt 23/4=-.6020, then -3 - sqrt23-4= -5.3979
the first one is + sorry
shouldn't the sqrt be 33?
probably lol. I'm sleepy. Make sure you divide by 4? the - one you just did looks a lil low. I think im going to go to bed, I have an 8am class tommorrow
ok good night thanks for helping me your awsome
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