Q.The number of real solutions of the equation $x^2 - 3|x| + 2 = 0$ are:

$|x+2|\le 9$ means $-9\le x+2\le 9$, which gives $x\in[-11,7]$.

By the definition of absolute value, $|x+2|\le 9$ is equivalent to $-9\le x+2\le 9$.

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Squaring the inequality (safe since both sides are $\geq0$) removes the modulus and reduces to a simple linear inequality.

We are given $|x-1|\geq|x-3|$. Since both sides are non-negative, we may square both sides without reversing the inequality:

$(x-1)^2\geq(x-3)^2$

Expanding, $x^2-2x+1\geq x^2-6x+9$.

The $x^2$ terms cancel: $-2x+1\geq-6x+9$. …

Solving $x^2-3|x|+2=0$ as a quadratic in $|x|$ gives two positive roots for $|x|$, each of which yields two real values of $x$, for 4 solutions total.

Let $y=|x|\ge 0$. Since $x^2=y^2$, the equation becomes:

$y^2 - 3y + 2 = 0 \implies (y-1)(y-2)=0 \implies y=1 \text{ or } y=2$

Both values are non-negative, so both are valid for $|x|$. …

Splitting the absolute value at $x=1$ and solving each case gives exactly two valid roots, $x=0$ and $x=1$.

Case $x\ge1$ (so $|x-1|=x-1$): $x^2+(x-1)=1 \Rightarrow x^2+x-2=0 \Rightarrow (x+2)(x-1)=0 \Rightarrow x=-2$ or $x=1$. Only $x=1$ satisfies $x\ge1$.

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$|x+2|\le 8$ means $-8\le x+2\le 8$; subtracting 2 throughout gives $-10\le x\le 6$.

$|x+2|\le 8$ unfolds to the compound inequality $-8 \le x+2 \le 8$.

Subtract 2 from all three parts: $-8-2 \le x \le 8-2$, i.e. $-10 \le x \le 6$.

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