The Square Root Property Recall: ππ = π Since this is a quadratic equation, we must set it equal to zero, factor, set each factor equal to zero, and solve. ππ = π ππ β π = π (π + π)(π β π) = π π+π=π π = βπ πβπ=π π=π Solving the equation mentally, we should ask ourselves, βwhat number when squared gives us π?β π squared is π and (βπ) squared is π. We know we should get π answers for a quadratic. When we want to solve this equation we can square root both sides, but we must keep in mind that there are two possible results: one positive and one negative. ππ = π βππ = ±βπ π = ±π Note: Recall that the symbol ± means βplus or minusβ. When taking the square root of both sides of an equation, we add ± to indicate both the positive AND negative solution. 1 Example 1: Solve using the square root property. a.) ππ = ππ b.) ππ = ππ c.) ππ + π = ππ Here we want to isolate the ππ so we can use the square root property. π₯ 2 + 1 = 26 Subtract π ππ = ππ Square root prop. βππ = ±βππ Simplify π = ±π Example 2: Solve using the square root property. a.) (π + π)π = π (π₯ + 1)2 = 8 β(π + π)π = ±βπ Square root prop. π + π = ±βπ Simplify π + π = ±πβπ Simplify π = βπ ± πβπ Subtract π 2 b.) πππ β π = ππ πππ β π = ππ Add π πππ = ππ Divide π ππ = π Square root prop. βππ = ±βπ Simplify π = ±βπ c.) πππ β ππ = ππ 3 The Square Root Property Practice Problems Solve each equation using the square root property. 1. ππ = πππ 2. ππ + π = ππ 3. πππ β π = ππ 4. (π + π)π = π 4
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