Understanding the Equation with Absolute Value

The problem asks for the number of real solutions to the equation $x^2 - 3 |x| + 2 = 0$. This equation involves the absolute value of $x$, denoted by $|x|$. Recall that the absolute value $|x|$ is defined as $x$ if $x \ge 0$ and $-x$ if $x < 0$.

A key property we can use is that $x^2 = |x|^2$ for any real number $x$. This is because squaring a number makes it non-negative, just like the absolute value operation. For example, $(-3)^2 = 9$ and $|-3|^2 = 3^2 = 9$.

Using this property, we can rewrite the given equation entirely in terms of $|x|$. The equation $x^2 - 3 |x| + 2 = 0$ becomes:

\begin{equation} |x|^2 - 3 |x| + 2 = 0 \end{equation}

Solving the Equation by Substitution

This equation looks like a standard quadratic equation if we consider $|x|$ as the variable. Let's make a substitution to make this clearer. Let $y = |x|$. Since $|x|$ is always non-negative, $y$ must be greater than or equal to 0 ($y \ge 0$).

Substituting $y$ into the equation, we get:

\begin{equation} y^2 - 3y + 2 = 0 \end{equation}

This is a quadratic equation in $y$. We can solve this by factoring, using the quadratic formula, or completing the square. Factoring is straightforward in this case.

We look for two numbers that multiply to $+2$ and add up to $-3$. These numbers are $-1$ and $-2$. So, we can factor the quadratic equation as:

\begin{equation} (y - 1)(y - 2) = 0 \end{equation}

This equation gives us two possible values for $y$:

• $y - 1 = 0 \implies y = 1$
• $y - 2 = 0 \implies y = 2$

Both of these values, $y=1$ and $y=2$, are non-negative, which is consistent with our substitution $y = |x|$.

Finding the Values of x

Now we need to substitute back $y = |x|$ to find the values of $x$.

Case 1: $y = 1$

Since $y = |x|$, this means $|x| = 1$. The equation $|x| = 1$ has two real solutions:

• $x = 1$ (because $|1| = 1$)
• $x = -1$ (because $|-1| = 1$)

Case 2: $y = 2$

Since $y = |x|$, this means $|x| = 2$. The equation $|x| = 2$ also has two real solutions:

• $x = 2$ (because $|2| = 2$)
• $x = -2$ (because $|-2| = 2$)

Counting the Real Solutions

The real solutions we found for the original equation $x^2 - 3 |x| + 2 = 0$ are $1, -1, 2,$ and $-2$. These are four distinct real numbers.

Let's verify these solutions in the original equation $x^2 - 3 |x| + 2 = 0$:

• For $x=1$: $(1)^2 - 3|1| + 2 = 1 - 3(1) + 2 = 1 - 3 + 2 = 0$. Correct.
• For $x=-1$: $(-1)^2 - 3|-1| + 2 = 1 - 3(1) + 2 = 1 - 3 + 2 = 0$. Correct.
• For $x=2$: $(2)^2 - 3|2| + 2 = 4 - 3(2) + 2 = 4 - 6 + 2 = 0$. Correct.
• For $x=-2$: $(-2)^2 - 3|-2| + 2 = 4 - 3(2) + 2 = 4 - 6 + 2 = 0$. Correct.

All four values are indeed real solutions.

Therefore, the number of real solutions is 4.

Revision Table: Solving Absolute Value Equations

Additional Information: Graphs and Solutions

We can also think about the number of real solutions graphically. The equation can be written as $x^2 + 2 = 3|x|$. We can plot the graphs of $y = x^2 + 2$ and $y = 3|x|$. The number of intersection points will give the number of real solutions.

• The graph of $y = x^2 + 2$ is a parabola opening upwards, with its vertex at $(0, 2)$. It is symmetric about the y-axis.
• The graph of $y = 3|x|$ consists of two lines: $y = 3x$ for $x \ge 0$ and $y = -3x$ for $x < 0$. This graph forms a V-shape with its vertex at $(0, 0)$.

Let's analyze the intersection points:

• For $x \ge 0$, the equation is $x^2 + 2 = 3x$, which is $x^2 - 3x + 2 = 0$. Factoring gives $(x-1)(x-2) = 0$, so $x=1$ or $x=2$. Both are $\ge 0$, so these are valid solutions.
• For $x < 0$, the equation is $x^2 + 2 = -3x$, which is $x^2 + 3x + 2 = 0$. Factoring gives $(x+1)(x+2) = 0$, so $x=-1$ or $x=-2$. Both are < 0, so these are valid solutions.

Thus, the intersection points occur at $x=-2, -1, 1, 2$. This confirms that there are 4 real solutions, corresponding to the intersection points of the two graphs.