Verify if the inequality is true. Discuss! 2x^2 + 5x<12
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:
x+4>0 and 2x-3<0
x>-4 and x<3/2
x belongs to the interval (-4,3/2).
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).
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.