The Mathematics 11 Competency Test Adding and Subtracting Square Roots We’ve already covered the addition and subtraction of numerical square roots in some detail. The procedure for adding and subtracting square roots which may contain algebraic expressions is more or less the same: • simplify the square root in each term of the expression • combine terms whose square roots are identical We will illustrate this strategy with a number of simple examples. Example 1: Simplify 5 y −6 y +8 y −2 y . solution: The square roots in all four of the terms in this expression are identical, all being just y . So, we just collect the “like terms:” 5 y − 6 y + 8 y − 2 y = y (5 − 6 + 8 − 2) = y (5) =5 y as the simplest form of the result. Example 2: Simplify 5 8 x 3 + 3 18 x 5 . solution: Neither of the two square roots occurring here are in simplest form. So, to start, we must simplify each term. Since 8x3 = 23x3 = 22•2•x2•x we can write 8x3 = 2 2 i 2 i x 2 i x = 22 x2 2x = 2x 2x Also David W. Sabo (2003) Adding and Subtracting Square Roots Page 1 of 6 18x5 = 2•32•(x2)2x and so ( ) 18 x 5 = 2 i 32 ⋅ x 2 2 (x ) i x = 32 2 2 2x = 3 x 2 2x Thus ( 5 8 x 3 + 3 18 x 5 = ( 5 ) 2 x 2 x ) + (3) (3x 2 2x ) = 10 x 2 x + 9 x 2 2 x ( = 10 x + 9 x 2 ) 2x You could leave the final answer as this last expression, or you could still do a bit of factoring to get, as the most simplified form 5 8 x 3 + 3 18 x 5 = x (10 + 9 x ) 2 x Example 3: Simplify 9 x 18 x 2 − 3 50 x 4 . solution: This example is very similar to Example 2 above, so you should use it as a practice problem. Try to solve it yourself before looking at our solution, given below. First, simplify the individual square roots where possible. Since 18x2 = 2•32•x2 we get 18 x 2 = 2 i 32 i x 2 = 32 x2 2 = 3x 2 and since 50x4 = 2•52•(x2)2 we get 50 x 4 = 2 i 52 i ( x 2 ) 2 = 52 (x ) 2 2 2 = 5x2 2 Thus David W. Sabo (2003) Adding and Subtracting Square Roots Page 2 of 6 ( ) ( 9 x 18 x 2 − 3 50 x 4 = ( 9 x ) 3 x 2 − ( 3 ) 5 x 2 = 27 x 2 2 − 15 x 2 = ( 27 x 2 − 15 x 2 ) 2 ) 2 2 = 12 x 2 2 as the final answer. 2x + Example 4: Simplify 8x + 18 x + 32 x . solution: The last three of the square roots here can be simplified slightly: 8x = 22 i 2 x = 2 2 x 18 x = 32 i 2 x = 3 2 x 32 x = 42 i 2 x = 4 2 x and Thus, 2x + 8x + 18 x + = 32 x = 2x + 2 2x + 3 2x + 4 2x 2 x (1 + 2 + 3 + 4 ) = 10 2x as the final result. Example 5: Simplify 5 pq 3 − 4 p pq + 6q 9 pq . solution: We have pq 3 = q 2 i pq = q pq 9 pq = 32 i pq = 3 pq and David W. Sabo (2003) Adding and Subtracting Square Roots Page 3 of 6 Thus ( 5 pq 3 − 4 p pq + 6q 9 pq = ( 5 ) q pq ) − 4p ( pq + ( 6q ) 3 pq ) = 5q pq − 4 p pq + 18q pq = pq ( 5q − 4 p + 18q ) = ( 23q − 4 p ) pq as the final answer. The method should be fairly obvious by now. Use the remaining three examples as practice problems. Try to do them yourself before looking at our solutions. Example 6: Simplify 3 12y 3 − 6 y 48 y . solution: First, we check to simplify each of the two square roots individually: 12y 3 = 22 i 3 i y 2 i y = 22 y2 3 y = 2y 3 y and 48 y = 42 i 3 y = 3y = 4 3y 42 Thus ( 3 12y 3 − 6 y 48 y = ( 3 ) 2y 3 y ) − ( 6y ) ( 4 3y ) = 6 y 3 y − 24 y 3 y = 3y ( 6y − 24y ) = −18y 3y as the final answer. David W. Sabo (2003) Adding and Subtracting Square Roots Page 4 of 6 Example 7: Simplify 7t 2 18s 5t − 3s 8s 3t 5 . solution: 18s 5t = 2 i 32 i ( s 2 ) i st = 3s 2 8s 3t 5 = 22 i 2 i s 2 i s i (t 2 ) i t = 2st 2 2 2st and 2 2st So ( ) ( 3s 7t 2 18s 5t − 3s 8s 3t 5 = 7t 2 2 2st = 21s 2t 2 = 2st ) − (3s ) ( 2st 2st − 6s 2t 2 ( 21s t 2 2 2 2st ) 2st ) − 6s 2t 2 = 15s 2t 2 2st as the final answer. Example 8: Simplify 3 p 8q 3 − 6q 32 p3 − 7 50q + 5 18 p . solution: Each square root must first be simplified: 8q 3 = 32 p3 = 22 i 2 i q 2 i q = 2q 2q 42 i 2 i p 2 i p = 4 p 2 p 50q = 52 i 2q = 5 2q 18 p = 32 i 2 p = 3 2 p and Therefore 3 p 8q 3 − 6q 32 p3 − 7 50q + 5 18 p ( = ( 3 p ) 2q 2q ) − ( 6q ) ( 4p ) ( 2 p − ( 7 ) 5 2q ) + (5 ) (3 2p ) = 6 pq 2q − 24 pq 2 p − 35 2q + 15 2 p David W. Sabo (2003) Adding and Subtracting Square Roots Page 5 of 6 = ( 6 pq − 35 ) 2q + (15 − 24 pq ) 2p This is as simple as we can get the expression. David W. Sabo (2003) Adding and Subtracting Square Roots Page 6 of 6
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