Question

How do you solve $9{x^2} + 3x - 2 \ge 0$?

Answer 1

Hint: We will have to solve for $x$ simply by solving the above equation, not just as an equation but with keeping inequality in mind, i.e., the addition or subtraction of variables and constants on both sides rather than simply transposing them. Then the inequality is quadratic in one variable, we will factorize the terms and find the value of $x$.

Complete step-by-step answer:
We have been given inequality $9{x^2} + 3x - 2 \ge 0$.
The steps to solving quadratic inequalities,
1. Factorize the quadratic equation by putting $a{x^2} + bx + c = 0$.
2. Determine the critical values i.e., values at which the expression becomes zero.
3. Plot these critical values and assign signs using the wavy curve method.
4. Plot a rough sketch or graph.
Now, factor the equation on the left side,
$ \Rightarrow 9{x^2} - 3x + 6x - 2 \ge 0$
Factor out 3x from the first group and 2 from the second group,
$ \Rightarrow 3x\left( {3x - 1} \right) + 2\left( {3x - 1} \right) \ge 0$
Now factor out $\left( {3x - 1} \right)$ from both groups,
$ \Rightarrow \left( {3x - 1} \right)\left( {3x + 2} \right) \ge 0$
As we know that, if $\left( {x - a} \right)\left( {x - b} \right) \ge 0$ and $a < b$. Then, $x \le a$ or $x \ge b$. So,
$ \Rightarrow 3x - 1 \ge 0$ and $3x + 2 \le 0$
Move the constant on the right side,
$ \Rightarrow 3x \ge 1$ and $3x \le - 2$
Divide both terms by 2,
$ \Rightarrow x \ge \dfrac{1}{3}$ and $x \le - \dfrac{2}{3}$

Hence, we can say that the value of real $x$ with given inequality is $x \in \left( { - \infty , - \dfrac{2}{3}} \right] \cup \left[ {\dfrac{1}{3},\infty } \right)$.

Note:
The students are likely to make mistakes in questions when finding the roots for > sign.
E.g.,
If $\left( {x - 2} \right)\left( {x - 5} \right) < 0$, then $2 < x < 5$.
If $\left( {x - 2} \right)\left( {x - 6} \right) > 0$, then $x < 2$ and $x > 6$.
Also, keep in mind the number of factors will never exceed the highest power of the polynomial. Also, students can always check the factors are correct or not by substituting the values in the polynomial, the polynomial turns out to be zero when we substitute the values.