Question

If r is a constant such that \(|x^2-4x-13| = r\) has exactly three distinct real roots, then the value of r is

• A. 15
• B. 18
• C. 17
• D. 21

Answer 1

The Correct Option is C

Given the equation \( |x^2 - 4x - 13| = r \), we need to find the value of \( r \) such that the equation has exactly three distinct real roots combined.

Step 1: Solve for the equation \( x^2 - 4x - 13 = r \)

The equation becomes: \[ x^2 - 4x - 13 - r = 0 \] Using the discriminant of a quadratic equation \( ax^2 + bx + c \), for real roots, the condition is \( b^2 - 4ac \geq 0 \). For this equation: \[ b^2 - 4ac = 16 - 4(1)(-13 - r) \] Simplifying: \[ 16 + 52 + 4r \geq 0 \] \[ 68 + 4r \geq 0 \] \[ r \leq -17 \]

Step 2: Solve for the equation \( x^2 - 4x - 13 = -r \)

The equation becomes: \[ x^2 - 4x - 13 + r = 0 \] Again, using the discriminant: \[ b^2 - 4ac = 16 - 4(1)(-13 + r) \] Simplifying: \[ 16 + 52 - 4r \geq 0 \] \[ 68 - 4r \geq 0 \] \[ r \leq 17 \]

Step 3: Determining the possible value of \( r \)

For the total of three real roots, the first equation will provide real roots when \( r \leq -17 \), and the second equation will provide real roots when \( r \leq 17 \). The only possible value for \( r \) that satisfies the conditions of exactly three real roots is: \[ r = 17 \]

Final Answer:

The correct option is \( \boxed{(C): 17} \).

Answer 2

Find the value of \( r \) for which the equation \[ |x^2 - 4x - 13| = r \] has exactly three different real roots.

Step 1: Complete the Square

The expression inside the absolute value is: \[ x^2 - 4x - 13 = (x - 2)^2 - 17 \] So the equation becomes: \[ |(x - 2)^2 - 17| = r \]

Step 2: Solve the Two Cases

Case 1: \[ (x - 2)^2 - 17 = r \Rightarrow (x - 2)^2 = r + 17 \] Case 2: \[ (x - 2)^2 - 17 = -r \Rightarrow (x - 2)^2 = -r + 17 \]

Step 3: Counting Distinct Real Roots

Each equation yields:

• Two real roots if the right-hand side is positive, i.e. square roots are real and distinct
• One real root if the right-hand side is zero
• No real root if negative (since square root of a negative number is not real)

 

To get exactly 3 real roots, one of these equations must yield 1 real root and the other must yield 2 real roots. That happens when: \[ \text{One RHS} = 0 \quad \text{and the other} > 0 \]

Step 4: Plug in Condition

Set either: \[ r + 17 = 0 \quad \text{(not valid, since } r = -17 \text{ gives negative RHS in other equation)} \] or \[ -r + 17 = 0 \Rightarrow r = 17 \] Then:

• First equation: \((x - 2)^2 = r + 17 = 34\) → 2 real roots
• Second equation: \((x - 2)^2 = -r + 17 = 0\) → 1 real root

 

✅ Final Answer: \(\boxed{17}\)