Solve $x^2-|5x-3|-x<2,\ \ x\in \mathbb{R} $
I tried $x^2-|5x-3|-x<2$ ,
case $1$ , $x^2-(5x-3)-x<2,\ x\geq 0 \\ x^2-6x+1<0 \\ 3-2\sqrt2 < 3+2\sqrt2 \\ 0.17<x<5.8\\ $
$x^2-(5x-3)-x<2$ ,
case $2$ , $x^2+(5x-3)-x<2,\ x< 0 \\ x^2+4x-5<0 \\ -5 < x< 1\\ $
The region common is $3-2\sqrt2<x<1$
But the book gives answer $-5<x<3+2\sqrt2$ . I am confused.
$\endgroup$2 Answers
$\begingroup$Case 1 is not for $x>0$ but for $5x-3>0\implies x>\frac{3}{5}$
So for case 1 you have $\frac{3}{5}<x<3+\sqrt{2}$ (since $\frac{3}{5}>3-\sqrt{2}$
For case 2 you have $x<\frac{3}{5}$, so $-5<x<\frac{3}{5}$
So the general solution is $-5<x<3+\sqrt{2}$
$\endgroup$ 1 $\begingroup$You can see why you need to discard the two solutions if you rearrange the equation to
$x^2-x-2<|5x-3|$
and draw them both.
When you solve for $x^2-x-2<5x-3$ you must only take into account that $x>\frac 35$
Similarily for $x^2-x-2<-5x+3$ you must only take into account that $x<\frac 35$
This is what makes the V-shape for the absolute value.
Then you only want the solutions where the y-values of the parabola are less than the y-values of the absolute value.
See Click on "Show Lines" box to see the two discarded solutions at point A and B.
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