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What is the number of real solutions? -11x^2 = x + 11

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first set equal to zero second find discriminate (b^2 -4ac) third, if discriminate >0 : 2 real solutions if discriminate =0: 1 real solution if discriminate <0 : no real solutions
would the equation be -11x^2 - x - 11 = 0 ?
yes or you can get rid of the neg --> 11x^2 +x+11 = 0

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Okay, now what would I do?
find Discriminant
Compare your quadratic equation with \(ax^2+bx+c=0\) find a,b,c then calculate \(b^2-4ac=...?\)
don't I already have a, b, and c so would I just substitute them into b^2 - 4ac ?
what u got, a=..? b=...?c=..?
doesn't a = 11, b = 1, and c = 11 ?
thats correct, now as @dumbcow said, u just need to find whether b^2-4ac is positive or negative...
so if I plug it in would it be like this .... \[1 ^{2} - 4(11)(11)\]
yes, go on.
okay so \[1^{2} = 1 and -4(11) = -44(11) = -484\] Then you would subtract -484 from 1 which gives you -483, right ?
yes, so how many real solutions ?
since -483 is less than 0 would it be no real solutions?
absolutely correct :)
Okay thank you so much for both of your helps!
welcome ^_^

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