Question

How do you solve the inequality \[16{x^2} + 8x + 1 > 0\] and write your answer in interval notation?

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 this using factoring or quadratic formula.

Complete step-by-step solution:
We have \[16{x^2} + 8x + 1 > 0\]
We cannot split the middle term, so we use a quadratic formula.
\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]. Consider \[16{x^2} + 8x + 1 = 0\]
Where \[a = 16,b = 8\] and \[c = 1\].
\[\Rightarrow x = \dfrac{{ - 8 \pm \sqrt {{8^2} - 4(16)(1)} }}{{2(16)}}\]
\[\Rightarrow x = \dfrac{{ - 8 \pm \sqrt {64 - 64} }}{{32}}\]
\[\Rightarrow x = \dfrac{{ - 8 \pm \sqrt 0 }}{{32}}\]
\[\Rightarrow x = \dfrac{{ - 8}}{{32}}\]
Thus we have two roots,
\[ \Rightarrow x = - \dfrac{1}{4}\] and \[x = - \dfrac{1}{4}\]
Thus we have
\[ \Rightarrow x = - 0.25\] and \[x = - 0.25\]
Thus we have \[ - \dfrac{1}{4} < x < - \dfrac{1}{4}\] or \[ - 0.25 < x < - 0.25\].
Lets put \[x = - 0.25\] or \[x = - \dfrac{1}{4}\] in \[16{x^2} + 8x + 1 > 0\]
\[\Rightarrow 16{\left( { - \dfrac{1}{4}} \right)^2} + 8\left( { - \dfrac{1}{4}} \right) + 1 > 0\]
\[\Rightarrow 16\left( {\dfrac{1}{{16}}} \right) - 8\left( {\dfrac{1}{4}} \right) + 1 > 0\]
\[
 \Rightarrow 1 - 2 + 1 > 0 \\
 \Rightarrow 0 > 0 \\
 \]
Which is not true. So the inequality satisfies for all the real values of ‘x’ except at \[x = - \dfrac{1}{4}\]
Let’s check by taking \[x = 1\] which is greater than \[ - \dfrac{1}{4}\]
\[
 \Rightarrow 16{(1)^2} + 8(1) + 1 > 0 \\
 \Rightarrow 16 + 8 + 1 > 0 \\
  \Rightarrow 25 > 0 \\
 \]
Which is true.
Let’s take \[x = - 1\] which is less than \[ - \dfrac{1}{4}\].
\[
\Rightarrow 16{( - 1)^2} + 8( - 1) + 1 > 0 \\
\Rightarrow 16 - 8 + 1 > 0 \\
\Rightarrow 9 > 0 \\
 \]
Which is true.
Hence the solution of \[16{x^2} + 8x + 1 > 0\] is \[\left( { - \infty , - \dfrac{1}{4}} \right) \cup \left( { - \dfrac{1}{4},\infty } \right)\].


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.