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### High School Mathematics2.4 Problems on Square Root - III

 Example: Solve 2(x-6)2 + 7 = 19 - (x-6)2. Solution: Let x-6 = a, the the given equation becomes, 2(a)2 + 7 = 19 - (a)2 2a2 + 7 - 19 + a2 = 0 3a2 - 12 = 0 a2 = 12/3 a2 = 4 Taking square root on both sides a = ±2 But a = x - 6 x - 6 = 2 or x - 6 = -2 x = 6 + 2 or x = -2 + 6 x = 8 or x = 4 Verification: When x = 8 Þ 2(x - 6)2 = 19 - (x - 6)2 2(8 - 6)2 = 19 - (8 - 6)2 2(2)2 = 19 - (2)2 2*4 + 7 = 19 - 4 8 + 7 = 19 - 4 15 = 15 \Both sides are equal, so the given equation is satisfied by x = 8. When x = 4Þ 2(x - 6)2 = 19 - (x - 6)2 2(4 - 6)2 = 19 - (4 - 6)2 2(-2)2 = 19 - (-2)2 2*4 + 7 = 19 - 4 8 + 7 = 19 - 4 15 = 15 \Both sides are equal, so the given equation is satisfied by x = 4. Directions: Solve the following problems. Also write at least ten examples of your own.
 Q 1: Solve 7 (x-3)2 - 8 = 40 - 5(x-3)2.2 and 63 and 71 and 5 Q 2: Solve 8(x+5)2 = 288.1 and -81 and -61 and -11 Q 3: Solve 9(x+5)2 = 576.4 and -143 and -163 and -13 Q 4: Solve 3(x+2)2 - 48 = 32 - 2(x+2)2.3 and -74 and -82 and -6 Question 5: This question is available to subscribers only! Question 6: This question is available to subscribers only!