Explanation Given: The equation 1 x 2 − 4 + 2 x − 2 = 1 x . Property used: Zero-Factor property : If a b = 0 , then either a = 0 or b = 0 or both, where a and b are real numbers. Addition property of equality : If a = b , then a + c = b + c , where a , b , and c are real numbers. Addition property , P Q + R Q = P + R Q , for all mathematical expressions P , Q , and R , with Q ≠ 0 . Multiplication property of equality : If a = b , then a c = b c , where a , b , and c are real numbers. Calculation: Factors of the denominator of the equation, x 2 − 4 = ( x ) 2 − ( 2 ) 2 = ( x + 2 ) ( x − 2 ) [ Difference of two squares ] Thus, the given equation becomes, 1 ( x + 2 ) ( x − 2 ) + 2 x − 2 = 1 x Each of these terms can be divided into x ( x + 2 ) ( x − 2 ) , so x ( x + 2 ) ( x − 2 ) is a common denominator. Multiply both sides of the equation by x ( x + 2 ) ( x − 2 ) . x ( x + 2 ) ( x − 2 ) ( 1 ( x + 2 ) ( x − 2 ) + 2 x − 2 ) = x ( x + 2 ) ( x − 2 ) ( 1 x ) x ( x + 2 ) ( x − 2 ) ( 1 ( x + 2 ) ( x − 2 ) + 2 x − 2 ) = ( x + 2 ) ( x − 2 ) Apply Distributive property, x ( x + 2 ) ( x − 2 ) ( 1 (

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Chapter R.4, Problem 4YT

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To solve: The equation 1x2−4+2x−2=1x by factoring or by using the quadratic formula.

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Chapter R.4, Problem 3YT

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The equation 1 x 2 − 4 + 2 x − 2 = 1 x by factoring or by using the quadratic formula .

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To solve: The equation 1x2−4+2x−2=1x by factoring or by using the quadratic formula.

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Chapter R Solutions