Example

3(r+2s)=2t-4
a⋅(b-2)=3b
3(r+2s)=2t-4
a⋅(b-2)=3b

Question

$$( { x }^{ 2 } -4x+4)(x-3) \geq 0FINDINEQUALITY$$

Answer

x>=3

Solution


Simplify  \(0FINDINEQUALITY\)  to  \(0\).
\[({x}^{2}-4x+4)(x-3)\ge 0\]
Rewrite \({x}^{2}-4x+4\) in the form \({a}^{2}-2ab+{b}^{2}\), where \(a=x\) and \(b=2\).
\[({x}^{2}-2(x)(2)+{2}^{2})(x-3)\ge 0\]
Use Square of Difference: \({(a-b)}^{2}={a}^{2}-2ab+{b}^{2}\).
\[{(x-2)}^{2}(x-3)\ge 0\]
Solve for \(x\).
\[x=2,3\]
From the values of \(x\) above, we have these 3 intervals to test.
\[\begin{aligned}&x\le 2\\&2\le x\le 3\\&x\ge 3\end{aligned}\]
Pick a test point for each interval.
For the interval \(x\le 2\):
Let's pick \(x=0\). Then, \(({0}^{2}-4\times 0+4)(0-3)\ge 0FINDINEQUALITY\).After simplifying, we get \(-12\ge 0\), which is
false
.
Drop this interval.
.
For the interval \(2\le x\le 3\):
Let's pick \(x=\frac{5}{2}\). Then, \(({(\frac{5}{2})}^{2}-4\times \frac{5}{2}+4)(\frac{5}{2}-3)\ge 0FINDINEQUALITY\).After simplifying, we get \(-0.125\ge 0\), which is
false
.
Drop this interval.
.
For the interval \(x\ge 3\):
Let's pick \(x=4\). Then, \(({4}^{2}-4\times 4+4)(4-3)\ge 0FINDINEQUALITY\).After simplifying, we get \(4\ge 0\), which is
true
.
Keep this interval.
.
Therefore,
\[x\ge 3\]
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