Question

How do you solve and write the following in interval notation:
$2{{x}^{2}}+5x-12\le 0$?

Answer 1

For solving the given inequality, we have to factor the quadratic polynomial in the LHS. For this, we can use the middle term splitting method according to which, the middle term $5x$ has to be split into two terms such that the product of the two terms is equal to the product of the first term $2{{x}^{2}}$ and the third term $-12$, that is equal to $-24{{x}^{2}}$. From the factored polynomial, we can finally write the solution for the given inequality in the interval notation.

Complete step-by-step answer:
The inequality given in the above question is
$\Rightarrow 2{{x}^{2}}+5x-12\le 0$
For solving the above inequality, we first need to factor the quadratic polynomial written on the left hand side. So we consider the quadratic polynomial as
$\Rightarrow p\left( x \right)=2{{x}^{2}}+5x-12$
Let us split the above polynomial using the middle term splitting method. We know that according to this method, we have to split the middle term as a sum of two terms such that the product of the two terms is equal to the first and the third term of the polynomial. In the above polynomial ,the first terms is equal to $2{{x}^{2}}$ and the third terms is equal to $-12$, which makes the product equal to $-24{{x}^{2}}$. Thus, we split the middle term $5x$ as $5x=8x-3x$ to get
$\Rightarrow p\left( x \right)=2{{x}^{2}}+8x-3x-12$
Now, taking $2x$ common from the first two terms, and $-3$ common from the last two terms, we get
$\Rightarrow p\left( x \right)=2x\left( x+4 \right)-3\left( x+4 \right)$
Now, we take the factor $\left( x+4 \right)$ common to get
$\Rightarrow p\left( x \right)=\left( x+4 \right)\left( 2x-3 \right)$
Putting this factored form of the polynomial in the given inequality, we get
$\Rightarrow \left( x+4 \right)\left( 2x-3 \right)\le 0$
Dividing both the sides by $2$ we get

& \Rightarrow \dfrac{\left( x+4 \right)\left( 2x-3 \right)}{2}\le \dfrac{0}{2} \\\ & \Rightarrow \left( x+4 \right)\dfrac{\left( 2x-3 \right)}{2}\le 0 \\\ & \Rightarrow \left( x+4 \right)\left( x-\dfrac{3}{2} \right)\le 0 \\\ \end{aligned}$$ On solving the above inequality, we get $\Rightarrow -4\le x\le \dfrac{3}{2}$ According to the question, we need to write the above solution in the interval notation. Thus, we get $$\Rightarrow x\in \left[ -4,\dfrac{3}{2} \right]$$ Hence, we have finally obtained the solution of the given inequality in the interval notation as $$x\in \left[ -4,\dfrac{3}{2} \right]$$. **Note:** Do not forget to write the solution in the interval notation, as written in the given question. Also, do not forget the equal to sign under the less than sign in the given inequality. It is due to this reason that the interval of the solution is closed from both the ends.