New SAT Math Workbook

Then square both sides to eliminate the radical sign

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: C) (D) (E) 93 ±9 3 33 ±3 3 ±9 5. none of these www.petersons.com Concepts of Algebra—Signed Numbers and Equations 125 5. EQUATIONS CONTAINING RADICALS In solving equations containing radicals, it is important to get the radical alone on one side of the equation. Then square both sides to eliminate the radical sign. Solve the resulting equation. Remember that all solutions to radical equations must be checked, as squaring both sides may sometimes result in extraneous roots. In squaring each side of an equation, do not make the mistake of simply squaring each term. The entire side of the equation must be multiplied by itself. Example: x –3 = 4 Solution: x – 3 = 16 x = 19 Checking, we have 16 = 4, which is true. Example: x – 3 = –4 Solution: x – 3 = 16 x = 19 Checking, we have 16 = –4, which is not true, since the radical sign means the principal, or positive, square root only. is 4, not –4; therefore, this equation has no solution. Example: x2 – 7 + 1 = x Solution: First get the radical alone on one side, then square. x2 – 7 = x – 1 x2 – 7 = x2 – 2x + 1 – 7 = – 2x + 1 2x = 8 x=4 Checking, we have 9 + 1 = 4 3 + 1 = 4, which is true. www.petersons.com 126 Chapter 8 Exercise 5 Work out each problem. Circle the letter that appears before your answer. 1. Solve for y: 2 y + 11 = 15 (A) 4 (B) 2 (C) 8 (D) 1 (E) no solution Solve for x: 4 2 x – 1 = 12 (A) 18.5 (B) 4 (C) 10 (D) 5 (E) no solution Solve for x: (A) (B) (C) (D) (E) x 2 – 35 = 5 – x 4. Solve for y: 26 = 3 2 y + 8 (A) 6 (B) 18 (C) 3 (D) –6 (E) no solution Solve for x: (A) (B) (C) (D) (E) 2x =4 5 2. 5. 3. 10 20 30 40 no solution 6 –6 3 –3 no solution www.petersons.com Concepts of Algebra—Signed Numbers and Equations 127 RETEST Work out each problem. Circle the letter that appears before your answer. 1. When –5 is subtracted from the sum of –3 and +7, the result is (A) +15 (B) –1 (C) –9 (D) +9 (E) +1 The product of – 2 (–4)(+12) – 6 is (A) 2 (B) –2 (C) 4 (D) –4 (E) –12 When the sum...
View Full Document

Ask a homework question - tutors are online