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??

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- Mar 22nd 2009, 10:39 PMscubasteve94Discriminant
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??

- Mar 22nd 2009, 10:59 PMADARSH
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" (Giggle)

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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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