Question 155·Medium·Nonlinear Functions

Factor $x^2 - 5x + 4$ and find the $x$-values where it is zero. Between and outside these values, is the expression positive or negative?

Remember that $|A| = A$ if $A \ge 0$ and $|A| = -A$ if $A < 0$. Apply this to $A = x^2 - 5x + 4$ on the intervals where it is positive and negative, and write separate equations for each case.

After solving the equations from each case, make sure each solution actually lies in the interval you assumed for that case (and satisfies $x \ge 0$). Then count how many distinct real solutions you have.

In Desmos, enter y = |x^2 - 5x + 4| on one line and y = x on another line to see both graphs on the same axes.

Since the absolute value is always nonnegative, focus on the part of the graph where $x \ge 0$. Adjust the viewing window if needed so you can clearly see where the line and the absolute value curve intersect on the right side of the $y$-axis.

Use the Desmos intersection tool (tap/click where the graphs cross) to mark each intersection point between the two graphs. The number of intersection points in the region $x \ge 0$ is the number of real solutions to the equation.

The left side is an absolute value, so it is always nonnegative.

That means the right side, $x$, must also be nonnegative:

So we only need to consider $x \ge 0$.

Factor the expression inside the absolute value:

This is:

• Zero at $x = 1$ and $x = 4$.
• Positive when $x < 1$ or $x > 4$.
• Negative when $1 < x < 4$.

Combining this with $x \ge 0$, we have three regions to consider:

• $0 \le x \le 1$ (expression $\ge 0$)
• $1 < x < 4$ (expression $< 0$)
• $x \ge 4$ (expression $\ge 0$).

In the regions where $x^2 - 5x + 4 \ge 0$ (that is, $0 \le x \le 1$ or $x \ge 4$), the absolute value does nothing:

So the equation becomes

Rearrange:

Solve using the quadratic formula:

Approximate to check the intervals:

• $3 - \sqrt{5} \approx 3 - 2.236 = 0.764$, which is between $0$ and $1$.
• $3 + \sqrt{5} \approx 3 + 2.236 = 5.236$, which is greater than $4$.

Both values lie in the regions where we assumed $x^2 - 5x + 4 \ge 0$, so both are valid solutions from this case.

In the region $1 < x < 4$, the expression $x^2 - 5x + 4$ is negative, so

Now the equation becomes

Rearrange:

Multiply by $-1$ to simplify:

Factor:

so $x = 2$.

Since $2$ is in the interval $1 < x < 4$, this value is a valid solution from this case.

From the two cases, we found:

• Two valid solutions from Case 1: $x = 3 - \sqrt{5}$ and $x = 3 + \sqrt{5}$.
• One valid solution from Case 2: $x = 2$.

All three solutions are real and satisfy the original equation.

Therefore, the equation has 3 real solutions, so the correct choice is C) Three.