Key Concepts and Formulas

• Definition of Absolute Value: $|x| = x$ if $x \ge 0$, and $|x| = -x$ if $x < 0$.
• Solving Quadratic Equations: For a quadratic equation of the form $ax^2 + bx + c = 0$, we can find the roots by factoring, completing the square, or using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
• Casework: When dealing with absolute values, we must consider different cases based on the sign of the expression inside the absolute value.

Step-by-Step Solution

Step 1: Rewrite the equation and introduce a substitution

We are given the equation $5 + |2x - 1| = 2x(2x - 2)$. To simplify the algebra, let $t = 2x$. The equation becomes: $5 + |t - 1| = t(t - 2)$ This substitution helps to reduce the complexity of the expressions involved.

Step 2: Analyze Case 1: $2x - 1 \ge 0$ (or $t - 1 \ge 0$)

• Condition on $x$: $2x - 1 \ge 0 \Rightarrow 2x \ge 1 \Rightarrow x \ge \frac{1}{2}$.
• Condition on $t$: Since $t = 2x$, $t \ge 1$.
• Simplify the absolute value: If $t \ge 1$, then $|t - 1| = t - 1$.
• Substitute into the equation: $5 + (t - 1) = t(t - 2)$ $4 + t = t^2 - 2t$

Step 3: Solve the quadratic equation for Case 1

Rearrange the equation into standard quadratic form: $t^2 - 3t - 4 = 0$ Factor the quadratic: $(t - 4)(t + 1) = 0$ This gives us two possible solutions for $t$: $t = 4$ and $t = -1$.

Step 4: Verify the solutions for $t$ in Case 1

We must check if the solutions satisfy the condition $t \ge 1$.

• If $t = 4$, then $t \ge 1$ is satisfied.
• If $t = -1$, then $t \ge 1$ is not satisfied, so $t = -1$ is an extraneous solution.

Step 5: Find the solution for $x$ in Case 1

Since $t = 4$ is a valid solution, we have $2x = 4$, which means $x = 2$. Also, we verify that $x=2$ satisfies $x \ge \frac{1}{2}$.

Step 6: Analyze Case 2: $2x - 1 < 0$ (or $t - 1 < 0$)

• Condition on $x$: $2x - 1 < 0 \Rightarrow 2x < 1 \Rightarrow x < \frac{1}{2}$.
• Condition on $t$: Since $t = 2x$, $t < 1$.
• Simplify the absolute value: If $t < 1$, then $|t - 1| = -(t - 1) = 1 - t$.
• Substitute into the equation: $5 + (1 - t) = t(t - 2)$ $6 - t = t^2 - 2t$

Step 7: Solve the quadratic equation for Case 2

Rearrange the equation into standard quadratic form: $t^2 - t - 6 = 0$ Factor the quadratic: $(t - 3)(t + 2) = 0$ This gives us two possible solutions for $t$: $t = 3$ and $t = -2$.

Step 8: Verify the solutions for $t$ in Case 2

We must check if the solutions satisfy the condition $t < 1$.

• If $t = 3$, then $t < 1$ is not satisfied, so $t = 3$ is an extraneous solution.
• If $t = -2$, then $t < 1$ is satisfied.

Step 9: Find the solution for $x$ in Case 2

Since $t = -2$ is a valid solution, we have $2x = -2$, which means $x = -1$. Also, we verify that $x=-1$ satisfies $x < \frac{1}{2}$.

Step 10: Count the number of real roots

From Case 1, we have $x = 2$. From Case 2, we have $x = -1$. Therefore, there are two distinct real roots.

Common Mistakes & Tips

• Forgetting to check the conditions: Always verify that the solutions obtained in each case satisfy the initial conditions for that case. Failing to do so can lead to extraneous solutions.
• Incorrectly handling the absolute value: Ensure you correctly apply the definition of absolute value based on the sign of the expression inside it.
• Algebraic errors: Be careful with algebraic manipulations, especially when rearranging and factoring quadratic equations.

Summary

We solved the equation $5 + |2x - 1| = 2x(2x - 2)$ by considering two cases based on the sign of $2x - 1$. In each case, we substituted $t=2x$ to simplify the resulting quadratic equation. After solving for $t$, we checked if the solutions satisfied the conditions for each case. Finally, we converted the valid $t$ values back to $x$ values, obtaining two real roots: $x=2$ and $x=-1$.

Final Answer The final answer is \boxed{2}, which corresponds to option (A).