Solution

Use the facts:

• $|A|=b\ (b\ge 0) \Rightarrow A=b\ \text{or}\ A=-b$
• $|A|<b\ (b>0) \Rightarrow -b<A<b$
• $|A|>b\ (b\ge 0) \Rightarrow A>b\ \text{or}\ A<-b$

(i) $|x^2+1|=5$

Second gives no real solution.

(ii) $|x^2+5x+4|=0$

Absolute value is zero only if the expression is zero:

(iii) $|x^2-6x+8|=4$

First:

Second:

(iv) $|3x^2-7x+2|=x^2-x+1$

Since $x^2-x+1>0$ for all real $x$, we can use cases:

Case 1:

Case 2:

So $x=\frac{1}{2}$ or $x=\frac{3}{2}$.

(v) $|x^2-4|<5$

Since $x^2\ge 0$, we only need $x^2<9$:

(vi) $|x^2-3x+2|>4$

First:

Roots of $x^2-3x-2=0$ are:

Since the parabola opens upward, $>0$ outside the roots:

Second:

(vii) $|x^2-5x+6|\le x+2$

Since $|\cdot|\ge 0$, we must have $x+2\ge 0\Rightarrow x\ge -2$.

Let $A=x^2-5x+6=(x-2)(x-3)$.

• If $A\ge 0$ (i.e. $x\le 2$ or $x\ge 3$), then $|A|=A$:

Roots: $x=3\pm\sqrt{5}$, so solution is:

Intersecting with $x\le 2$ or $x\ge 3$ keeps:

• If $A<0$ (i.e. $2<x<3$), then $|A|=-A$:

This is always true, so all $2<x<3$ works.

Combine all parts:

(viii) $|2x^2-3x-5|<4$

Let $A=2x^2-3x-5$.

So:

1. $2x^2-3x-9<0$ has roots $x=-\frac{3}{2}$ and $x=3$, so:

2. $2x^2-3x-1>0$ has roots $x=\frac{3\pm\sqrt{17}}{4}$, so:

Intersect with $\left(-\frac{3}{2},3\right)$: