Question:
The quadratic expression having zero's equal to the non-integral roots of \( ||3x-4|-6|=5 \) is:
The quadratic expression having zero's equal to the non-integral roots of \( ||3x-4|-6|=5 \) is:
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Always simplify nested absolute value problems by systematically removing the outer layers, which reduces complex equations into simpler linear segments.
Updated On: Jun 9, 2026
- \( x^2 + \frac{2}{3}x - \frac{35}{9} \)
- \( x^2 + \frac{4}{5}x + \frac{17}{25} \)
- \( x^2 - \frac{5}{7}x + \frac{12}{49} \)
- \( x^2 + \frac{1}{2}x + \frac{15}{4} \)
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The Correct Option is A
Solution and Explanation
Concept:
The problem requires solving an absolute value equation with nested modulus signs. Recall that \( |f(x)| = a \) (where \( a > 0 \)) branches into two equations: \( f(x) = a \) or \( f(x) = -a \). We will solve the outermost layer, then the inner layer, identify the non-integral roots, and construct the quadratic.
Step 1: Resolve the outer absolute value.
Given \( ||3x-4|-6|=5 \): Case I: \( |3x-4| - 6 = 5 \Rightarrow |3x-4| = 11 \) Case II: \( |3x-4| - 6 = -5 \Rightarrow |3x-4| = 1 \)
Step 2: Solve each case to find the roots.
For Case I (\( |3x-4| = 11 \)): \( 3x-4 = 11 \Rightarrow 3x = 15 \Rightarrow x = 5 \) (Integer) \( 3x-4 = -11 \Rightarrow 3x = -7 \Rightarrow x = -7/3 \) (Non-integral) For Case II (\( |3x-4| = 1 \)): \( 3x-4 = 1 \Rightarrow 3x = 5 \Rightarrow x = 5/3 \) (Non-integral) \( 3x-4 = -1 \Rightarrow 3x = 3 \Rightarrow x = 1 \) (Integer)
Step 3: Form the quadratic expression.
The non-integral roots are \(\alpha = -7/3\) and \(\beta = 5/3\). Sum of roots: \( \alpha + \beta = -7/3 + 5/3 = -2/3 \). Product of roots: \( \alpha \beta = (-7/3) \times (5/3) = -35/9 \). The quadratic is \( x^2 - (\text{Sum})x + (\text{Product}) = 0 \): \[ x^2 - (-2/3)x + (-35/9) = x^2 + \frac{2}{3}x - \frac{35}{9} \] x^2 + 23x - 359
Step 1: Resolve the outer absolute value.
Given \( ||3x-4|-6|=5 \): Case I: \( |3x-4| - 6 = 5 \Rightarrow |3x-4| = 11 \) Case II: \( |3x-4| - 6 = -5 \Rightarrow |3x-4| = 1 \)
Step 2: Solve each case to find the roots.
For Case I (\( |3x-4| = 11 \)): \( 3x-4 = 11 \Rightarrow 3x = 15 \Rightarrow x = 5 \) (Integer) \( 3x-4 = -11 \Rightarrow 3x = -7 \Rightarrow x = -7/3 \) (Non-integral) For Case II (\( |3x-4| = 1 \)): \( 3x-4 = 1 \Rightarrow 3x = 5 \Rightarrow x = 5/3 \) (Non-integral) \( 3x-4 = -1 \Rightarrow 3x = 3 \Rightarrow x = 1 \) (Integer)
Step 3: Form the quadratic expression.
The non-integral roots are \(\alpha = -7/3\) and \(\beta = 5/3\). Sum of roots: \( \alpha + \beta = -7/3 + 5/3 = -2/3 \). Product of roots: \( \alpha \beta = (-7/3) \times (5/3) = -35/9 \). The quadratic is \( x^2 - (\text{Sum})x + (\text{Product}) = 0 \): \[ x^2 - (-2/3)x + (-35/9) = x^2 + \frac{2}{3}x - \frac{35}{9} \] x^2 + 23x - 359
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