Question

How do you solve the inequality $9{x^2} - 6x + 1 \leqslant 0$?

Answer 1

An inequality compares two values, showing if one is less than, greater than, or simply not equal to another value. Here we need to solve for ‘x’ which is a variable. Solving the given inequality is very like solving equations and we do most of the same thing but we must pay attention to the direction of inequality$( \leqslant , > )$. We have a quadratic equation we can solve using factorization.

Complete step by step solution:
We have $9{x^2} - 6x + 1 \leqslant 0$
We can split the middle term, we have:

$$\Rightarrow 9{x^2} - 3x - 3x + 1 \leqslant 0 \\\ \Rightarrow 3x(3x - 1) - 1(3x - 1) \leqslant 0 \\\ \Rightarrow (3x - 1)(3x - 1) \leqslant 0 \\\$$

Thus we have $x \leqslant \dfrac{1}{3}$.
Since we have less than or equal to inequality sign, we need to be careful.
If we put a value $x < \dfrac{1}{3}$ in $9{x^2} - 6x + 1 \leqslant 0$ it doesn’t satisfy. That is
That is $x < 0.333$.
Put $x = 0.1$ in $9{x^2} - 6x + 1 \leqslant 0$

$$\Rightarrow 9{(0.1)^2} - 6(0.1) + 1 \leqslant 0 \\\ \Rightarrow 9{(0.1)^2} - 6(0.1) + 1 \leqslant 0 \\\ \Rightarrow 0.09 - 0.6 + 1 \leqslant 0 \\\ \Rightarrow 0.49 \leqslant 0 \\\$$

Which is a contradiction because 0.49 is not less than 0.
Hence the solution of $9{x^2} - 6x + 1 \leqslant 0$ is $x = \dfrac{1}{3}$

Note: We know that $a \ne b$ says that ‘a’ is not equal to ‘b’. $a > b$ means that ‘a’ is less than ‘b’. $a < b$ means that ‘a’ is greater than ‘b’. These two are known as strict inequality. $a \geqslant b$ means that ‘a’ is less than or equal to ‘b’. $a \leqslant b$ means that ‘a’ is greater than or equal to ‘b’.

The direction of inequality do not change in these cases:
i) Add or subtract a number from both sides.
ii) Multiply or divide both sides by a positive number.
iii) Simplify a side.

The direction of the inequality change in these cases:
i) Multiply or divide both sides by a negative number.
ii) Swapping left and right hand sides.