HW1-solutions - wei (jw35975) – HW1 – milburn –...

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Unformatted text preview: wei (jw35975) – HW1 – milburn – (54685) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Which, if any, of the following statements are true when a, b are real numbers? A. For all positive a and b . a − b √ a + √ b = √ a + √ b . B. For all a and b , radicalBig ( a + b ) 2 = | a + b | . C. For all positive a and b , √ a + b = √ a + √ b . 1. B and C only 2. none of them 3. B only correct 4. A only 5. all of them 6. A and C only 7. A and B only 8. C only Explanation: A. FALSE: by the known difference of squares factorization, x 2 − y 2 = ( x − y )( x + y ) . But if a, b are positive we can set x = √ a and y = √ b . Thus, after division, a − b √ a + √ b = √ a − √ b , contrary to the assertion. B. TRUE: we know that radicalBig ( x + y ) 2 = | x + y | , and since radicalbig ( · ) is always non-negative, the right hand side has to be non-negative. That’s why the absolute value sign is needed. C. FALSE: by the known product, ( x + y ) 2 = x 2 + 2 xy + y 2 . On the other hand, radicalBig ( x + y ) 2 = | x + y | , so if x + y > 0, x + y = radicalbig x 2 + 2 xy + y 2 . But if a, b are positive we can set x = √ a and y = √ b , in which case √ a + √ b = radicalBig a + 2 √ ab + b , contrary to the assertion. keywords: square root, properties of square root, PlaceUT, TrueFalse, T/F, 002 10.0 points Simplify the expression f ( x ) = 3 + 9 x − 4 1 + 15 parenleftBig x x 2 − 16 parenrightBig as much as possible. 1. f ( x ) = 3( x − 4) x + 16 2. f ( x ) = x + 4 x − 16 3. f ( x ) = 3( x + 4) 2 x + 16 wei (jw35975) – HW1 – milburn – (54685) 2 4. f ( x ) = 3( x + 4) x + 16 correct 5. f ( x ) = x − 4 x − 16 6. f ( x ) = x − 4 2 x − 16 Explanation: After bringing the numerator to a common denominator it becomes 3 x − 12 + 9 x − 4 = 3 x − 3 x − 4 . Similarly, after bringing the denominator to a common denominator and factoring it be- comes x 2 − 16 + 15 x x 2 − 16 = ( x − 1)( x + 16) x 2 − 16 . Consequently, f ( x ) = 3 + 9 x − 4 1 + 15 parenleftBig x x 2 − 16 parenrightBig = 3 x − 3 ( x − 1)( x + 16) parenleftBig x 2 − 16 x − 4 parenrightBig . On the other hand, x 2 − 16 = ( x + 4)( x − 4) . Thus, finally, we see that f ( x ) = 3( x + 4) x + 16 . 003 10.0 points Let f be the quadratic function defined by f ( x ) = 2 x 2 − 12 x − 14 . Use completing the square to find h so that f ( x ) = 2( x − h ) 2 + k for some value of k . 1. h = 12 2. h = − 6 3. h = 3 correct 4. h = − 3 5. h = 6 Explanation: Completing the square gives f ( x ) = 2 x 2 − 12 x − 14 = 2( x 2 − 6 x − 7) = 2( x 2 − 6 x + 9 − 7 − 9) ....
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This note was uploaded on 10/25/2011 for the course MATHEMATIC 408 taught by Professor Milburn during the Fall '11 term at University of Texas at Austin.

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HW1-solutions - wei (jw35975) – HW1 – milburn –...

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