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2.2 Nonlinear Equations in One Variable - Quadratic solving (factoring, quadratic formula, completing square)
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the fraction with numerator 4 x squared, and denominator x squared minus 9, end fraction, minus the fraction with numerator 2 x, and denominator x plus 3, end fraction, equals the fraction with numerator 1, and denominator x minus 3, end fraction

What value of x satisfies the equation above?

A.

negative 3

B.

negative one half

C.

one half

D.

3

What value of x satisfies the equation above?

Hard-difficulty · SAT Math · Nonlinear Equations in One Variable — Quadratic solving (factoring, quadratic formula, completing square). Read the question above, select your answer, and check the full explanation below to understand exactly why the correct choice works.

Answer explanation

Choice C is correct. Each fraction in the given equation can be expressed with the common denominator x squared, minus 9. Multiplying the fraction with numerator 2 x, and denominator x plus 3, end fraction by the fraction with numerator x minus 3, and denominator x minus 3, end fraction yields the fraction with numerator 2 x squared, minus 6, and denominator x squared minus 9, end fraction , and multiplying the fraction with numerator 1, and denominator x minus 3, end fraction by the fraction with numerator x plus 3, and denominator x plus 3, end fraction yields the fraction with numerator x plus 3, and denominator x squared minus 9, end fraction. Therefore, the given equation can be written as the fraction with numerator 4 x squared, and denominator x squared minus 9, minus, the fraction with numerator 2 x squared, minus 6 x, and denominator x squared minus 9, end fraction, equals, the fraction with numerator x plus 3, and denominator x squared minus 9, end fraction. Multiplying each fraction by the denominator results in the equation 4 x squared minus, open parenthesis, 2 x squared, minus 6 x, close parenthesis, equals, x plus 3, or 2 x squared, plus 6 x, equals, x plus 3. This equation can be solved by setting a quadratic expression equal to 0, then solving for x. Subtracting x plus 3 from both sides of this equation yields 2 x squared, plus 5 x, minus 3, equals 0. The expression 2 x squared, plus 5 x, minus 3 can be factored, resulting in the equation open parenthesis, 2 x minus 1, close parenthesis, times, open parenthesis, x plus 3, close parenthesis, equals 0. By the zero product property, 2 x minus 1, equals 0 or x plus 3, equals 0. To solve for x in 2 x minus 1, equals 0, 1 can be added to both sides of the equation, resulting in 2 x equals 1. Dividing both sides of this equation by 2 results in x equals one half. Solving for x in x plus 3, equals 0 yields x equals negative 3. However, this value of x would result in the second fraction of the original equation having a denominator of 0. Therefore, x equals negative 3 is an extraneous solution. Thus, the only value of x that satisfies the given equation is x equals one half.

Choice A is incorrect and may result from solving x plus 3, equals 0 but not realizing that this solution is extraneous because it would result in a denominator of 0 in the second fraction. Choice B is incorrect and may result from a sign error when solving 2 x minus 1, equals 0 for x. Choice D is incorrect and may result from a calculation error.

 

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