New SAT Math Workbook

This equation could also have been solved by

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Unformatted text preview: for x. 2x = 0 3x – 4 = 0 x=0 3x = 4 x= 4 3 4 . 3 The roots of the given quadratic are 0 and Example: x2 – 9 = 0 Solution: x2 = 9 x=±3 Remember there must be two roots. This equation could also have been solved by factoring x2 – 9 into (x + 3)(x – 3) and setting each factor equal to 0. Remember that the difference of two perfect squares can always be factored, with one factor being the sum of the two square roots and the second being the difference of the two square roots. Example: x2 – 8 = 0 Solution: Since 8 is not a perfect square, this cannot be solved by factoring. x2 = 8 x=± 8 Simplifying the radical, we have 4· 2 , or x = ±2 2 www.petersons.com 124 Chapter 8 Example: 16x2 – 25 = 0 Solution: Factoring, we have (4x – 5) (4x + 5) = 0 x=± 5 4 Setting each factor equal to 0, we have If we had solved without factoring, we would have found 16x2 = 25 x2 = 25 16 5 x=± 4 Example: x2 + 6 x + 8 = 0 Solution: (x + 2)(x + 4) = 0 If the last term of the trinomial is positive, both binomial factors must have the same sign, since the last two terms multiply to a positive product. If the middle term is also positive, both factors must be positive since they also add to a positive sum. Setting each factor equal to 0, we have x = –4 or x = –2 Example: x2 – 2x – 15 = 0 Solution: We are now looking for two numbers that multiply to –15; therefore they must have opposite signs. To give –2 as a middle coefficient, the numbers must be –5 and +3. (x – 5)(x + 3) = 0 This equation gives the roots 5 and –3. Exercise 4 Work out each problem. Circle the letter that appears before your answer. 1. Solve for x: x2 – 8x – 20 = 0 (A) 5 and –4 (B) 10 and –2 (C) –5 and 4 (D) –10 and –2 (E) –10 and 2 Solve for x: 25x2 – 4 = 0 (A) (B) (C) (D) (E) 4 4 and – 25 25 2 2 and – 5 5 2 only 5 2 – only 5 3. Solve for x: 6x2 – 42x = 0 (A) (B) (C) (D) (E) 7 only –7 only 0 only 7 and 0 –7 and 0 2. 4. Solve for x: x2 – 19x + 48 = 0 (A) 8 and 6 (B) 24 and 2 (C) –16 and –3 (D) 12 and 4 (E) none of these Solve for x: 3x2 = 81 (A) (B) (...
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