Question

Solve the following inequality:
$$(x^2-2x)(2x-2)-9{\displaystyle \frac{2x-2}{x^2-2x}}⩽0$$

Answer 1

Hint: We have to find the value of $$x$$ from the given expression of inequality $$(x^2-2x)(2x-2)-9{\displaystyle \frac{2x-2}{x^2-2x}}⩽0$$. We solve this question using the concept of solving linear equations of inequality. We will first simplify the given equation by taking $$(2x-2)$$ common from the left-hand side of the inequality. Then we factorise the obtained expression into its factors. And then by further solving we will find the range for the values of $$x$$.

Complete step by step answer:
Given, $$(x^2-2x)(2x-2)-{\displaystyle \frac{9\left(2x-2\right)}{x^2-2x}}⩽0$$
By taking $$(2x-2)$$ common from left-hand side of the inequality, we get
$$\Rightarrow (2x-2)\left[(x^2-2x)-{\displaystyle \frac{9}{x^2-2x}}\right]⩽0$$
On solving, we get
$$\Rightarrow (2x-2)\left[{\displaystyle \frac{{(x^2-2x)}^2-9}{(x^2-2x)}}\right]⩽0$$
On rewriting the above inequality, we get
$$\Rightarrow (2x-2)\left[{\displaystyle \frac{{(x^2-2x)}^2-{(3)}^2}{(x^2-2x)}}\right]⩽0$$
Simplifying the numerator using the formula: $$a^2-b^2=\left(a-b\right)\left(a+b\right)$$, we get
$$\Rightarrow (2x-2)\left[{\displaystyle \frac{\left(x^2-2x-3\right)\left(x^2-2x+3\right)}{(x^2-2x)}}\right]⩽0$$
Adding and subtracting in $$1$$ in $$\left(x^2-2x+3\right)$$, we get
$$\Rightarrow (2x-2)\left[{\displaystyle \frac{\left(x^2-2x-3\right)\left(x^2-2x+1+3-1\right)}{(x^2-2x)}}\right]⩽0$$
As $$\left(x^2-2x+1\right)={\left(x-1\right)}^2$$, we can write the above inequality as
$$\Rightarrow (2x-2)\left[{\displaystyle \frac{\left(x^2-2x-3\right)\left({\left(x-1\right)}^2+2\right)}{(x^2-2x)}}\right]⩽0$$
On further factorising and we get
$$\Rightarrow (2x-2)\left[{\displaystyle \frac{\left(x^2+x-3x-3\right)\left({\left(x-1\right)}^2+2\right)}{(x^2-2x)}}\right]⩽0$$
Taking $$x$$ common and on simplifying we get,
$$\Rightarrow (2x-2)\left[{\displaystyle \frac{\left(x+1\right)\left(x-3\right)\left({\left(x-1\right)}^2+2\right)}{(x^2-2x)}}\right]⩽0$$
As $${\left(x-1\right)}^2+2>0$$, we can divide both the sides by $$\left({\left(x-1\right)}^2+2\right)$$, therefore we get
$$\Rightarrow (2x-2){\displaystyle \frac{\left(x+1\right)\left(x-3\right)}{(x^2-2x)}}⩽0$$
Taking $$2$$ common from the numerator and $$x$$ from denominator and on simplifying we get
$$\Rightarrow {\displaystyle \frac{2(x-1)\left(x+1\right)\left(x-3\right)}{x(x-2)}}⩽0$$
Dividing both the sides by $$2$$, we get
$$\Rightarrow {\displaystyle \frac{(x-1)\left(x+1\right)\left(x-3\right)}{x(x-2)}}⩽0$$
Now we will put all the values of $$x$$ on number line for which equality holds,

Let’s take any value of $$x$$ greater than $$3$$, say $$x=4$$, we get the value of $${\displaystyle \frac{(x-1)\left(x+1\right)\left(x-3\right)}{x(x-2)}}$$ as
$$\Rightarrow {\displaystyle \frac{(4-1)\left(4+1\right)\left(4-3\right)}{4(4-2)}}={\displaystyle \frac{15}{8}}>0$$
On taking any value greater than $$3$$, we get the value of inequality greater than zero.
Now taking any number between $$3$$ and $$2$$, say $$2.5$$, we get
$$\Rightarrow {\displaystyle \frac{(2.5-1)\left(2.5+1\right)\left(2.5-3\right)}{\left(2.5\right)(2.5-2)}}=-2.1<0$$
Therefore, on taking any value between $$3$$ and $$2$$, we will get the value of inequality less than zero.
Now taking any number between $$2$$ and $$1$$, say $$1.5$$, we get
$$\Rightarrow {\displaystyle \frac{(1.5-1)\left(1.5+1\right)\left(1.5-3\right)}{\left(1.5\right)(1.5-2)}}=2.5>0$$
Therefore, on taking any value between $$2$$ and $$1$$, we will get the value of inequality greater than zero.
Now, taking any number between $$1$$ and $$0$$, say $$0.5$$, we get
$$\Rightarrow {\displaystyle \frac{(0.5-1)\left(0.5+1\right)\left(0.5-3\right)}{\left(0.5\right)(0.5-2)}}=-2.5<0$$
Therefore, on taking any value between $$1$$ and $$0$$, we will get the value of inequality less than zero.
Now, taking any number between $$0$$ and $$-1$$, say $$-0.5$$, we get
$$\Rightarrow {\displaystyle \frac{(-0.5-1)\left(-0.5+1\right)\left(-0.5-3\right)}{\left(-0.5\right)(-0.5-2)}}=2.1>0$$
Therefore, on taking any value between $$0$$ and $$-1$$, we will get the value of inequality greater than zero.
Now, taking any number less than $$-1$$, say $$-2$$, we get
$$\Rightarrow {\displaystyle \frac{(-2-1)\left(-2+1\right)\left(-2-3\right)}{\left(-2\right)(-2-2)}}=-{\displaystyle \frac{15}{8}}<0$$
Therefore, on taking any number less than $$-1$$, we will get the value of inequality less than zero.
Also, at $$x=-1,0,1,2,3$$ equality holds.
Therefore, the interval of $$x$$ for which $$(x^2-2x)(2x-2)-9{\displaystyle \frac{2x-2}{x^2-2x}}⩽0$$ is when
$$\Rightarrow x\in (-\mathrm{\infty },-1]\cup \left[0,1\right]\cup \left[2,3\right]$$

Note:
The solution range of inequality gives us each and every value of $$x$$ which satisfies the equation. Here, one point to note is that square bracket $$\left[\right]$$ states that the end elements of the range will satisfy the given expression whereas the round bracket $$\left(\right)$$ states that the end elements will not satisfy the given expression.