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Show that $5|x|+x(x-2) \geq 0$ for every real number $x.$
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Case 1:-

let , $x > 0 $ so , $|x|=x$

So , given expression ,

$5\left | x \right |+x^{2}-2x$

= $5x+x^{2}-2x$

=$x^{2}+3x$

Now with any value of $x>0$ this expression $x^{2}+3x>0$ .

Case 2:-

let , $x < 0 $ so , $|x|=-x$

So , given expression ,

$5\left | x \right |+x^{2}-2x$

= $5(-x)+x^{2}-2x$

=$x^{2}-7x$

Now for all value of $x<0$ , $x^{2}$>0 and $(-7x)$ is also positive as $x<0$ .

So , $x^{2}-7x$ >0 for all $x<0$.

Case 3:-

Let $x=0$ so , $|x|=0$

So , given expression ,

$5\left | x \right |+x^{2}-2x$

= $5*0+0^{2}-2*0$

=$0$

So, Given expression $5\left | x \right |+x(x-2) \geq 0$ for all real $x$ .

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