Question

How do you solve $1\le \dfrac{3{{x}^{2}}-7x+8}{{{x}^{2}}+1}\le 2$?

Answer 1

In the given question, we have asked to solve the given inequality for the value of ‘x’. This is a quadratic inequality in one variable as there is only one variable in an equation. Solving quadratic inequality is the same as solving general quadratic equation but there is one more rule i.e. changing the sign of inequality if we multiply both sides by a negative number. To solve this inequality for a given variable ‘$x$’, we have to use the quadratic formula or splitting the middle term method and undo the mathematical operations such as addition, subtraction, multiplication and division that have been done to the variables. In this way we will get our required answer.

Complete step by step solution:
We have given that,
$1\le \dfrac{3{{x}^{2}}-7x+8}{{{x}^{2}}+1}\le 2$
As the given denominator i.e. ${{x}^{2}}+1$ is positive.
Multiply the given inequality by${{x}^{2}}+1$, we obtained
$1\left( {{x}^{2}}+1 \right)\le \dfrac{3{{x}^{2}}-7x+8}{{{x}^{2}}+1}\times \left( {{x}^{2}}+1 \right)\le 2\left( {{x}^{2}}+1 \right)$
Simplifying the above given inequality, we have
$\Rightarrow {{x}^{2}}+1\le 3{{x}^{2}}-7x+8\le 2{{x}^{2}}+2$
Now,
Taking the first inequality i.e.
$\Rightarrow {{x}^{2}}+1\le 3{{x}^{2}}-7x+8$
Subtracting 1 from both the sides of inequality, we get
$\Rightarrow {{x}^{2}}\le 3{{x}^{2}}-7x+8-1$
$\Rightarrow {{x}^{2}}\le 3{{x}^{2}}-7x+7$
Subtracting ${{x}^{2}}$from both the sides of the equation, we obtained
$\Rightarrow {{x}^{2}}-{{x}^{2}}\le 3{{x}^{2}}-7x+7-{{x}^{2}}$
Combining the like terms, we get
$\Rightarrow 0\le 2{{x}^{2}}-7x+7$
Rewritten the inequality as,
$\Rightarrow 2{{x}^{2}}-7x+7\ge 0$
Finding the discriminant,
D = ${{b}^{2}}-4ac={{\left( -7 \right)}^{2}}-4\times 2\times 7=49-56=-7$
Finding the value of ‘x’,
$x=\dfrac{\pm b-\sqrt{{{b}^{2}}-4ac}}{2a}=\dfrac{\pm \left( -7 \right)-\sqrt{{{\left( -7 \right)}^{2}}-4\times 2\times 7}}{2}$
Simplifying the above, we get
$x=\dfrac{\pm \left( -7 \right)-\sqrt{-7}}{2}$
The discriminant has negative value.
Hence, there is no real solution for this given inequality.
Since D < 0 for the given inequality.
So, the equation is true for all the value of ‘x’.
Now,
Taking the second inequality,
$\Rightarrow 3{{x}^{2}}-7x+8\le 2{{x}^{2}}+2$
Subtracting 2 from both the sides of inequality, we get
$\Rightarrow 3{{x}^{2}}-7x+8-2\le 2{{x}^{2}}$
$\Rightarrow 3{{x}^{2}}-7x+6\le 2{{x}^{2}}$
Subtracting $2{{x}^{2}}$from both the sides of inequality, we get
$\Rightarrow 3{{x}^{2}}-7x+6-2{{x}^{2}}\le 2{{x}^{2}}-2{{x}^{2}}$
Simplifying the above, we get
$\Rightarrow {{x}^{2}}-7x+6\le 0$
Finding the value of ‘x’,
We have,
$\Rightarrow {{x}^{2}}-7x+6\le 0$
Splitting the middle term, we get
$\Rightarrow {{x}^{2}}-6x-x+6\le 0$
Taking out the common factors, we get
$\Rightarrow x\left( x-6 \right)-1\left( x-6 \right)\le 0$
$\Rightarrow \left( x-1 \right)\left( x-6 \right)\le 0$
Therefore,
$x=6\ and\ 1$

Hence, $x\in \left[ 1,6 \right]$
This, it is the required answer.

Note: Use addition or subtraction properties of inequality to gather variable terms on one side of the inequality and constant on the other side of the inequality. The important thing to recollect about any inequality is that the ‘equals’ sign represents a balance. Use the multiplication or division properties of equality to form the coefficient of the variable term equivalent to 1. In solving inequality there is one sign changing rule i.e. sign of the inequality will be changed if we multiply both the sides of the inequality by negative number. This is the type of question where only mathematical operations such as addition, subtraction, multiplication and division is used.