Question

How do you solve the inequality \[2{x^2} + 9x + 3 \leqslant 0\]?

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 this using factoring or quadratic formula.

Complete step-by-step solution:
We have \[2{x^2} + 9x + 3 \leqslant 0\]
We cannot split the middle term, so we use a quadratic formula.
\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]. Consider \[2{x^2} + 9x + 3 = 0\]
Where \[a = 2,b = 9\] and \[c = 3\].
\[x = \dfrac{{ - 9 \pm \sqrt {{9^2} - 4(2)(3)} }}{{2(2)}}\]
\[x = \dfrac{{ - 9 \pm \sqrt {81 - 24} }}{4}\]
\[ \Rightarrow x = \dfrac{{ - 9 \pm \sqrt {57} }}{4}\]
Thus we have two roots,
\[ \Rightarrow x = \dfrac{{ - 9 + \sqrt {57} }}{4}\] and \[x = \dfrac{{ - 9 - \sqrt {57} }}{4}\]
Thus we have
\[ \Rightarrow x = \dfrac{{ - 9 + 7.55}}{4}\] and \[x = \dfrac{{ - 9 - 7.55}}{4}\]
\[ \Rightarrow x = \dfrac{{ - 1.45}}{4}\] and \[x = \dfrac{{ - 16.55}}{4}\]
\[ \Rightarrow x = - 0.3625\] and \[x = - 4.1375\]
Thus we have \[ - 4.1375 \leqslant x \leqslant - 0.3625\] or \[\dfrac{{ - 9 - \sqrt {57} }}{4} \leqslant x \leqslant \dfrac{{ - 9 + \sqrt {57} }}{4}\].
Let’s take a value of lies in \[ - 4.1375 \leqslant x \leqslant - 0.3625\]
Let’s put \[x = - 3\] in \[2{x^2} + 9x + 3 \leqslant 0\],
\[
\Rightarrow 2{( - 1)^2} + 9( - 1) + 3 \leqslant 0 \\
\Rightarrow 2 - 9 + 3 \leqslant 0 \\
\Rightarrow 2 - 9 + 3 \leqslant 0 \\
 \Rightarrow - 4 \leqslant 0 \\
 \]
Which is correct.
(If we take the value of ‘x’ lies outside of\[ - 4.1375 \leqslant x \leqslant - 0.3625\] the inequality does not satisfy.)
That is put \[x = - 5\] in \[2{x^2} + 9x + 3 \leqslant 0\],
\[
\Rightarrow 2{( - 5)^2} + 9( - 5) + 3 \leqslant 0 \\
\Rightarrow 50 - 45 + 3 \leqslant 0 \\
\Rightarrow 8 \leqslant 0 \\
 \]
Which is incorrect.)
Hence the solution of \[2{x^2} + 9x + 3 \leqslant 0\] is \[ - 4.1375 \leqslant x \leqslant - 0.3625\] or \[\dfrac{{ - 9 - \sqrt {57} }}{4} \leqslant x \leqslant \dfrac{{ - 9 + \sqrt {57} }}{4}\].
In the interval form \[[ - 4.1375, - 0.3625]\].

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.