Find the values of x for which 5x – 2 + 3x^2 is less than zero.
I know you have to let ∆ ≤ 0 but then im unsure how to do it??
for a quadratic equation
ax^2 +bx+c = 0
discriminant = b^2 - 4ac
If discriminant =< 0
Then roots are unreal
Discriminant for your equation is 25 +(4*3*2) = 25+24 = 49
Dont "assume the discriminant if its already given"
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For your question
5x – 2 + 3x^2 < 0
you need to find the solution set of x for which this inequality holds
3x^2 + 5x - 2 <0
3x^2 + 6x - x - 2 < 0
3x(x+2) - 1(x+2) <0
(3x-1)(x+2)<0
(x - 1/3) (x+2) <0
When x> 1/3
then
both terms are positive so this is not correct
when -2<x <1/3
First term is negative and 2nd is positive
Hence its negative thus its a solution set
similarly when x<-2
LHS is positive thus its not a solution set
Thus your answer is -2<x<1/3
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