Verify if the inequality is true. Discuss! 2x^2 + 5x<12

giorgiana1976 | Student

First, we have to subtract 12, both sides of the equation (it's easier to verify if an expression is negative or positive).

2x^2 + 5x - 12 < 0

Now, it would be much more easier to factorize the expression. In order to do so, we'll find out the roots of the equation 2x^2 + 5x - 12 = 0

x1 = [-5+sqrt(25+96)]/4

x1 = (-5+11)/4

x1 = 6/4

x1 = 3/2

x2 = (-5-11)/4

x2 = -16/4

x2 = -4

Now, we'll factorize the expression 2x^2 + 5x - 12:

2x^2 + 5x - 12 = 2(x+4)(x-3/2)

2x^2 + 5x - 12 = (x+4)(2x-3)

Now, we'll discuss when (x+4)(2x-3)<0.

For a product to be negative, the factors of the product have to have opposite signs. From here, it results 2 cases:

First case:

x+4>0 and 2x-3<0

x>-4 and x<3/2

x belongs to the interval (-4,3/2).

Second case:

x+4<0 and 2x-3>0

x<-4 and x>3/2

It is obvious that there are no real numbers to satisfy this condition, simultaneously.

So, the inequality is verified for x belongs to the interval (-4,3/2).

neela | Student

To solve 2x^2+5x<12.

Solution: This is a quadratic inequality which we solve by setting  f(x) = 2x^2+5x-12  and factorising and then finding the domain of x forwhich rhe range values of f(x) > 0

f(x) = 2x^2+8x-3x-12 = 2x(x+4)+3(x+4) = (x+4)(x-3).

Therefore the critical values of x for which f(x) = 0 are x+4 = 0 or x=-4 and x-3 = 0 or x=3.

Therefore f(x) = (x+4)(x-3) . Both factors are positive when x > 3.So for x>3 f(x) is positive.

When x < -4, both factos (x+4) and (x-3) are negative and f(x) = (x+4)(x+3) is positive.

When x lies between the critical values -4 and 3, Or -4<x<3  (x+4)(x-3) is negative. Or f(x) = 2x^2+5x-12 < 0. Or  2x^2+5x < 12 for x fo rwhich -4 <x <3.

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