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Zorluk: OrtaInequalities and Absolute Value Equations

What is the sum of all real solutions to the equation ∣x2−10∣=3x|x^2 - 10| = 3x?

Cevap: 7

Cevap

The sum of all real solutions to the equation is 7.
To solve ∣x2−10∣=3x|x^2 - 10| = 3x, note that 3x≥03x \ge 0 (so x≥0x \ge 0). Splitting into two algebraic cases gives x2−3x−10=0x^2 - 3x - 10 = 0 (yielding x=5x = 5 and extraneous x=−2x = -2) and x2+3x−10=0x^2 + 3x - 10 = 0 (yielding x=2x = 2 and extraneous x=−5x = -5). The valid real solutions are x=5x = 5 and x=2x = 2, whose sum is 7.

Adım Adım Çözüm

1
Determine domain restriction based on the absolute value definition.
x≥0x \ge 0
Because an absolute value expression cannot be negative, ∣x2−10∣=3x|x^2 - 10| = 3x requires 3x≥03x \ge 0, which means x≥0x \ge 0.
2
Solve Case 1: x2−10=3xx^2 - 10 = 3x.
x=5x = 5
Rearranging yields x2−3x−10=0x^2 - 3x - 10 = 0, which factors as (x−5)(x+2)=0(x - 5)(x + 2) = 0. The roots are x=5x = 5 and x=−2x = -2. Reject x=−2x = -2 because x≥0x \ge 0.
3
Solve Case 2: −(x2−10)=3x-(x^2 - 10) = 3x.
x=2x = 2
Rearranging yields x2+3x−10=0x^2 + 3x - 10 = 0, which factors as (x+5)(x−2)=0(x + 5)(x - 2) = 0. The roots are x=2x = 2 and x=−5x = -5. Reject x=−5x = -5 because x≥0x \ge 0.
4
Sum all valid real solutions.
7
The valid real solutions are x=5x = 5 and x=2x = 2. Their sum is 5+2=75 + 2 = 7.

Anahtar Kavram

Solving absolute value equations with a variable expression on one side requires verifying non-negativity constraints to eliminate extraneous roots.
Tahmini Süre:1m 30s
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Konular

Inequalities and Absolute Value Equations

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