Question

How do you solve $2{{x}^{2}}+3x>0$ ?

Answer 1

Hint: In this question, we have to find the value of x. The inequation is in the form of a quadratic; therefore, we will apply the basic mathematical rules to get the required solution. We will first take x common on the left-hand side of the inequation, and thus get two separate inequalities. Then, we will solve them individually and make the necessary calculations to get the required range of x.

Complete step by step answer:
According to the question, we have to find the value of x from an inequation.
The inequation given to us is $2{{x}^{2}}+3x>0$ ----------------- (1)
First, we will take common x from the left-hand side of the equation (1), we get
$x(2x+3)>0$
Thus, on further simplification, we get four new inequations, that is
$x>0$ , and ---------- (2)
$(2x+3)>0$ -------- (3)
$x<0$ , and ---------- (4)
$(2x+3)<0$ -------- (5)
We have taken both the equation less than zero in the above case, because we know that if the 2 polynomials are less than zero and are multiplied to each other, thus we get the result as greater than zero.
So, we will now first solve inequation (2), which is
$\Rightarrow x>0$
Thus, from the above inequation, we get that x will take all those values which are greater than 0, that is x takes 1, 2, 3… and so on .
Thus we get the range for x when $x>0$ is $\left( 0,\infty \right)$ ------- (6)
So, we will now first solve inequation (3), which is
$\Rightarrow (2x+3)>0$
Now, we will subtract 3 on both sides in the above inequation, e get
$\Rightarrow 2x+3-3>0-3$
As we know, the same terms with opposite signs cancel out each other, thus we get
$\Rightarrow 2x>-3$
Now, we will divide 2 on both sides in the above inequation, we get
$\Rightarrow \dfrac{2}{2}x>\dfrac{-3}{2}$
On further simplification, we get
$\Rightarrow x>\dfrac{-3}{2}$
Thus, the range of x for equation $(2x+3)>0$ is $\left( \dfrac{-3}{2},0 \right)$ ------ (7)
So, we will now first solve inequation (4), which is
$\Rightarrow x<0$
Thus, from the above inequation, we get that x will take all those values which are lesser than 0, that is x takes …, -4, -3, -2, -1 .
Thus we get the range for x when $x<0$ is $\left( -\infty ,0 \right)$ ------- (8)
So, we will now first solve inequation (5), which is
$\Rightarrow (2x+3)<0$
Now, we will subtract 3 on both sides in the above inequation, we get
$\Rightarrow 2x+3-3<0-3$
As we know, the same terms with opposite signs cancel out each other, thus we get
$\Rightarrow 2x<-3$
Now, we will divide 2 on both sides in the above inequation, we get
$\Rightarrow \dfrac{2}{2}x<\dfrac{-3}{2}$
On further simplification, we get
$\Rightarrow x<\dfrac{-3}{2}$
Thus, the range of x for equation $(2x+3)<0$ is $\left( -\infty ,\dfrac{-3}{2} \right)$ ------ (9)
Thus, from the value of equation (6), (7), (8), and (9), we get the range as $\left( 0,\infty \right)$, $\left( \dfrac{-3}{2},0 \right)$ , $\left( -\infty ,0 \right)$ , and $\left( -\infty ,\dfrac{-3}{2} \right)$ .
Therefore, for the inequation $2{{x}^{2}}-25x=0$ , we get the value of $x=\left( -\dfrac{3}{2},0 \right)\cup \left( 0,\infty \right)$ .

Note:
While solving this problem, do step-by-step calculations to avoid mathematical errors. Do not forget, that the given problem is a type of inequation and not equation, that is we have to find the range of x, which is the accurate answer.