Question

How do you solve the quadratic inequality ${{x}^{2}}-9x+18>0$ ?

Answer 1

Hint: To solve the given inequality we will first find the roots of the corresponding equality. Now let us say $\alpha $ and $\beta $ are the roots of the equation. Now we will check for three intervals which are $\left( -\infty ,\alpha \right),\left( \alpha ,\beta \right),\left( \beta ,\infty \right)$ by taking one value from each interval and hence check if the equation holds in the interval.

Complete step-by-step solution:
Now we are given with the quadratic inequality ${{x}^{2}}-9x+18>0$ .
Now to solve the roots of the inequality we will first find the roots of the corresponding quadratic equation.
Hence let us consider the quadratic equation ${{x}^{2}}-9x+18=0$ .
Now we know that ffor any equation of the form $a{{x}^{2}}+bx+c$ the solution of the equation is given by $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .
Now comparing the equation ${{x}^{2}}-9x+18=0$ with $a{{x}^{2}}+bx+c$ we get a = 1, b = - 9 and
c = 18.
Hence the roots of the quadratic equation are given by
$\begin{align}
  & \Rightarrow x=\dfrac{-\left( -9 \right)\pm \sqrt{{{\left( -9 \right)}^{2}}-4\left( 1 \right)\left( 18 \right)}}{2\left( 1 \right)} \\
 & \Rightarrow x=\dfrac{9\pm \sqrt{81-72}}{2} \\
 & \Rightarrow x=\dfrac{9\pm \sqrt{9}}{2} \\
 & \Rightarrow x=\dfrac{9-3}{2} \\
\end{align}$
Hence the value of x is $x=\dfrac{9+3}{2}$ or $x=\dfrac{9-3}{2}$
Hence x = 6 and x = 3.
Hence the roots of the equation are 3 to 6.
Now we will consider three intervals.
$\left( -\infty ,3 \right),\left( 3,6 \right)$ and $\left( 6,\infty \right)$ .
Now let us take value from each interval and check if the inequality holds..
From x = - 4 we have
$\Rightarrow {{\left( -4 \right)}^{2}}-9\left( -4 \right)+18=70>0$
Hence the inequality holds for $\left( -\infty ,3 \right)$
Now let us check for $\left( 3,6 \right)$ and hence let us check for x = 4.
Hence we get $16-9\left( 4 \right)+18=-2<0$
Hence the inequality does not holds for $\left( 3,6 \right)$
Now let us check the interval $\left( 6,\infty \right)$ .
Hence let us take the value 7.
Hence we get $49-9\left( 7 \right)+18=4>0$
Hence the inequality holds. Hence the inequality is true for $\left( 6,\infty \right)$
Hence the solution set for ${{x}^{2}}-9x+18>0$ is $\left( -\infty ,3 \right)\cup \left( 3,\infty \right)$.

Note: Now note that there is no need to check the solution for all intervals. Just check if the inequality for the equation between the roots of the equation. If it holds then the solution is just the interval between the roots. If the equation doesn’t hold between the roots then the solution set is $\left( -\infty ,\alpha \right)\cup \left( \beta ,\infty \right)$ where $\alpha $ and $\beta $ are the roots of the equation.