Algebra 1 Chapter 10: Square Root Equations Name: ______________________________ 1. As we learned when working through the Pythagorean Theorem, squares and square roots are inverse operations, they βundoβ each other. When solving an equation with a square, we take the square root on both sides. When solving an equation with a square root, we square each side. Solve the following radical equations by squaring both sides. a. π=π b. π=π c. π = ππ 2. Sometimes there is more under the radical than just the variable. But because the square and square root βundoβ each other, the value under the radical doesnβt change. For instance, if I square ππ + π I will just get 3b + 4. Solve the following radical equations while showing work. a. ππ = π b. ππ = ππ c. d. ππ β π = π e. ππ + π = π f. ππ = π 1 = βππ β π 3. The equations in problems #1 and 2 all had isolated radicals (square roots that were by themselves on one side of the equation). If the radical isnβt isolated, you need to do that first, then square both sides. Solve these one while showing work. Be sure to get the radical by itself first, then square both sides. a. π + π = ππ b. π β π = ππ c. π + ππ = π d. ππ + π = ππ e. ππ + π = ππ f. 3 β π = βπ 4. Sometimes there are radicals on both sides of the equations. Simply start by squaring both sides, then solving the resulting equation. a. ππ + π = ππ β π b. ππ = πβπ d. π + ππ = π β π e. π + π = ππ β ππ c. ππ β π = ππ + ππ f. ππ + π = ππ β π New Topic that also involves radicals: 5. Recall collecting like terms. Simplify the following a. ππ + ππ b. πππ β ππ c. 9π + ππ β ππ + ππ 6. If I tell you that π = π and π = π, then the above expressions could be written as the expressions shown below. The mathematics of simplifying these is the same as for #5, add the coefficients. Simplify these problems then check them on a calculator. a. π π+π π ππ π β π π b. c. 9 π+π πβπ π+π π 7. Simplify the following radical expressions by collecting like terms. a. π+π π b. ππ π β π π c. π π+ π d. ππ + π π β π ππ e. π β π π + π π β π π f. πππ ππ + ππ ππ
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