Question

The quadratic expression having zero's equal to the non-integral roots of \( ||3x-4|-6|=5 \) is:

• A. \( x^2 + \frac{2}{3}x - \frac{35}{9} \)
• B. \( x^2 + \frac{4}{5}x + \frac{17}{25} \)
• C. \( x^2 - \frac{5}{7}x + \frac{12}{49} \)
• D. \( x^2 + \frac{1}{2}x + \frac{15}{4} \)

Hint:

Always simplify nested absolute value problems by systematically removing the outer layers, which reduces complex equations into simpler linear segments.

Answer 1

The Correct Option is A

Step 1: Peel the outer modulus.
Given $\big|\,|3x-4| - 6\,\big| = 5$. The outer absolute value splits into $|3x-4| - 6 = 5$ or $|3x-4| - 6 = -5$.
Step 2: Get two inner equations.
These give $|3x-4| = 11$ (Case I) and $|3x-4| = 1$ (Case II).
Step 3: Solve Case I.
$3x-4 = 11 \Rightarrow x = 5$ (integer), and $3x-4 = -11 \Rightarrow x = -\tfrac{7}{3}$ (non-integer).
Step 4: Solve Case II.
$3x-4 = 1 \Rightarrow x = \tfrac{5}{3}$ (non-integer), and $3x-4 = -1 \Rightarrow x = 1$ (integer).
Step 5: Keep the non-integer roots.
The non-integral roots are $\alpha = -\tfrac{7}{3}$ and $\beta = \tfrac{5}{3}$. Their sum is $-\tfrac{7}{3} + \tfrac{5}{3} = -\tfrac{2}{3}$ and product is $-\tfrac{7}{3}\cdot\tfrac{5}{3} = -\tfrac{35}{9}$.
Step 6: Build the quadratic.
Using $x^2 - (\text{sum})x + (\text{product})$: \[ x^2 - \left(-\tfrac{2}{3}\right)x + \left(-\tfrac{35}{9}\right) = x^2 + \tfrac{2}{3}x - \tfrac{35}{9}, \] which is option (A).
\[ \boxed{x^2 + \tfrac{2}{3}x - \tfrac{35}{9}} \]