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Difficulty: HardRadical and Rational Equations

If xx is the real solution to the equation below, what is the value of xx?

2x2−30x+100x−5+140x+2=10\frac{2x^2 - 30x + 100}{x - 5} + \frac{140}{x + 2} = 10

Answer: 8

Answer

The correct answer is 8.
To solve the rational equation, we first state the domain restrictions as x≠5x \neq 5 and x≠−2x \neq -2 because these values make the denominators zero. Next, we factor the numerator of the first term: 2x2−30x+100=2(x2−15x+50)=2(x−5)(x−10)2x^2 - 30x + 100 = 2(x^2 - 15x + 50) = 2(x - 5)(x - 10). Since x≠5x \neq 5, we can cancel the common factor of x−5x - 5 from the numerator and denominator, which simplifies the first term to 2(x−10)=2x−202(x - 10) = 2x - 20. The equation can then be rewritten as 2x−20+140x+2=102x - 20 + \frac{140}{x + 2} = 10. Subtracting 10 from both sides yields 2x−30+140x+2=02x - 30 + \frac{140}{x + 2} = 0. Dividing the entire equation by 2 simplifies the coefficients, giving x−15+70x+2=0x - 15 + \frac{70}{x + 2} = 0. To clear the fraction, we multiply the entire equation by x+2x + 2, leading to (x−15)(x+2)+70=0(x - 15)(x + 2) + 70 = 0. Expanding the product gives x2−13x−30+70=0x^2 - 13x - 30 + 70 = 0, which simplifies to x2−13x+40=0x^2 - 13x + 40 = 0. Factoring this quadratic equation yields (x−5)(x−8)=0(x - 5)(x - 8) = 0, giving the potential solutions x=5x = 5 and x=8x = 8. However, x=5x = 5 is extraneous because it is restricted from the domain of the original rational expression. Therefore, the only real solution is 8.

Step-by-Step Solution

1
Find the values of xx that make any denominator equal to zero to identify domain restrictions.
x≠5x \neq 5 and x≠−2x \neq -2
Dividing by zero is undefined in real numbers, so these values are excluded from the domain.
2
Factor the numerator of the first term, 2x2−30x+1002x^2 - 30x + 100.
2(x−5)(x−10)2(x - 5)(x - 10)
Factoring allows us to check for common factors in the numerator and denominator.
3
Simplify the first rational term by dividing the numerator and denominator by the common factor x−5x - 5, under the assumption that x≠5x \neq 5.
2(x−10)=2x−202(x - 10) = 2x - 20
Reducing the expression simplifies the algebraic structure of the equation.
4
Substitute the simplified expression back into the original equation and subtract 10 from both sides.
2x−30+140x+2=02x - 30 + \frac{140}{x + 2} = 0
To consolidate the constant terms and prepare the equation for clearing the remaining fraction.
5
Divide the entire equation by 2.
x−15+70x+2=0x - 15 + \frac{70}{x + 2} = 0
Dividing by the common factor simplifies the coefficients before expanding.
6
Multiply the entire equation by x+2x + 2 to clear the denominator, then expand and combine like terms.
x2−13x+40=0x^2 - 13x + 40 = 0
This transforms the rational equation into a standard quadratic equation.
7
Factor the quadratic equation.
(x−5)(x−8)=0(x - 5)(x - 8) = 0
Factoring is the most direct method to find the roots of the quadratic equation.
8
Verify the candidate solutions against the domain restrictions identified in Step 1.
The candidate solution x=5x = 5 is extraneous because it is excluded from the domain of the original equation. The candidate solution x=8x = 8 is valid.
Extraneous solutions introduced during algebraic steps must be excluded from the final solution set.

Key Concept

Solving rational equations by simplifying terms, finding common denominators, and identifying extraneous solutions.
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Topics

Radical and Rational Equations

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