Question

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

Answer 1

Hint: 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.