Question

Difficulty: HardReal Numbers, Number Line, and Absolute Value

Let aa and bb be non-zero real numbers such that ∣a+b∣=3∣a−b∣|a + b| = 3|a - b|. What is the value of ∣a2+b2ab∣\left|\frac{a^2 + b^2}{ab}\right|?

  1. A
    32\frac{3}{2}
  2. B
    22
  3. 52\frac{5}{2}Answer
  4. D
    33
  5. E
    103\frac{10}{3}

Answer

The value of ∣a2+b2ab∣\left|\frac{a^2 + b^2}{ab}\right| is 52\frac{5}{2}.
Squaring both sides of the given equation ∣a+b∣=3∣a−b∣|a + b| = 3|a - b| eliminates the absolute values to give (a+b)2=9(a−b)2(a + b)^2 = 9(a - b)^2. Expanding and simplifying yields 2a2−5ab+2b2=02a^2 - 5ab + 2b^2 = 0, which factors as (2a−b)(a−2b)=0(2a - b)(a - 2b) = 0. Thus, b=2ab = 2a or a=2ba = 2b. In either case, substituting into ∣a2+b2ab∣\left|\frac{a^2 + b^2}{ab}\right| reduces the expression to 52\frac{5}{2}.

Step-by-Step Solution

1
Square both sides of the absolute value equality
(a+b)2=9(a−b)2(a + b)^2 = 9(a - b)^2
Since both sides of ∣a+b∣=3∣a−b∣|a + b| = 3|a - b| are non-negative real numbers, squaring preserves equality and eliminates the absolute value signs.
2
Expand both algebraic expressions and collect like terms
a2+2ab+b2=9a2−18ab+9b2  ⟹  8a2−20ab+8b2=0  ⟹  2a2−5ab+2b2=0a^2 + 2ab + b^2 = 9a^2 - 18ab + 9b^2 \implies 8a^2 - 20ab + 8b^2 = 0 \implies 2a^2 - 5ab + 2b^2 = 0
Expanding the binomial squares allows combining like terms to solve for the relationship between aa and bb.
3
Factor the quadratic expression in terms of aa and bb
(2a−b)(a−2b)=0  ⟹  b=2a or a=2b(2a - b)(a - 2b) = 0 \implies b = 2a \text{ or } a = 2b
Factoring determines the exact proportional relationship between aa and bb.
4
Substitute the relation into the target expression
\left|\frac{a^2 + (2a)^2}{a(2a)}\right| = \left|\frac{5a^2}{2a^2}\right| = \frac{5}{2}
Substituting b=2ab = 2a allows a2a^2 to cancel completely, leaving a constant numerical value.

Key Concept

Properties of Real Numbers and Absolute Value Equations
Estimated Time:2m 0s
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