MATH 1010-2: Quiz 9A Solution 1. (7 points) Solve the following equation using any method you choose. x2 + 2x + 5 = 0 Solution: By quadratic formula, x= −b ± √ b2 − 4ac 2a where a = 1, b = 2 and c = 5. Substitute the value of a, b and c into quadratic formula, we have √ −2 ± 22 − 4 · 1 · 5 x= 2·1 √ −2 ± −16 = 2 −2 ± 4i = 2 = −1 ± 2i In conclusion, the two solutions of the original equation are −1 + 2i and −1 − 2i. 2. (3 points) Solve the following equation using any method you choose. √ x− x−2=0 √ Solution: Since x = ( x)2 , the original equation can be rewritten as √ √ ( x)2 − x − 2 = 0 hence can be regarded as a standard quadratic equation in variable By factoring the trinomial, we have √ √ ( x − 2)( x + 1) = 0 or therefore √ x − 2 = 0 or √ x=2 or √ √ x+1=0 x = −1 take the square x=4 or x=1 Check the solutions, only x = 4 is solution of the original equation. √ x. MATH 1010-2: Quiz 9B Solution 1. (7 points) Solve the following equation using any method you choose. x2 − 2x + 5 = 0 Solution: By quadratic formula, x= −b ± √ b2 − 4ac 2a where a = 1, b = −2 and c = 5. Substitute the value of a, b and c into quadratic formula, we have p −(−2) ± (−2)2 − 4 · 1 · 5 x= 2·1 √ 2 ± −16 = 2 2 ± 4i = 2 = 1 ± 2i In conclusion, the two solutions of the original equation are 1 + 2i and 1 − 2i. 2. (3 points) Solve the following equation using any method you choose. √ x+ x−2=0 √ Solution: Since x = ( x)2 , the original equation can be rewritten as √ √ ( x)2 + x − 2 = 0 hence can be regarded as a standard quadratic equation in variable By factoring the trinomial, we have √ √ ( x + 2)( x − 1) = 0 or therefore √ x+2=0 √ or x = −2 or √ x−1=0 √ x=1 take the square x=4 or x=1 Check the solutions, only x = 1 is solution of the original equation. √ x.

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# MATH 1010-2: Quiz 9A

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