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4.7 Radical Equations and Problem Solving Power Rule • When solving radical equations, we use a new principle called the power rule. – The Power Rule states that if a = b, then an = bn and vice versa. • This also applies to radicals, specifically: n n a b. – If a = b, then Example 1 • If a radical with a variable is already isolated in an equation, we will raise each side of the equation to a power that matches the radical index. • Solve the radical equations below. x5 6 3 j 2 3 Extraneous Solutions • As was the case with rational equations, it was necessary to check answers to be sure they verified the original equation. • With radical equations, it is also necessary for two primary reasons: – With an even index, we cannot have a negative radicand (i.e. we cannot take the square root of a negative) – Unless specified otherwise, we will only calculate the principal root of any radical. • Example 2: Solve solutions. x 3 7and check your Solving polynomial inequalities • Rewrite the polynomial so that all terms are on one side and zero on the other. • Factor the polynomial. We are interested in when factors are either pos. or neg., so we must know when the factor equals zero. • The values of x for which the factors equal zero are the boundary points, which we place on the number line. • The intervals around the boundary points must be tested to find on which interval(s) will the polynomial be positive/negative. Solve : (x – 3)(x + 1)(x – 6) < 0 • To solve this inequality we observe that 0 is already on one side and the polynomial is factored already on the other side. • The 3 boundary values are x = 3,-1,6 • They create 4 intervals: (,1), (1,3), (3,6), (6, ) • Pick a number in each interval to test the sign of that interval. If the polynomial is negative there then the interval is in the solution set. • Solution set: Solve: x3 +3x2 ≥ 10x 1. To solve, first we must rewrite the inequality so all terms are on one side and 0 on the other, then factor. 2. x(x-2)(x+5) ≥ 0 3. Boundary points: 0, 2, -5 4. Solution set: Solving rational inequalities • VERY similar to solving polynomial inequalites EXCEPT if the denominator equals zero, there is a domain restriction. The function COULD change signs on either side of that point. • Step 1: Rewrite the inequality so all terms are on one side and zero on the other. • Step 2: Factor both numerator & denominator to find boundary values for regions to check when function becomes positive or negative. And do as before ! Solve the following inequalities: 1) 2) 3) x 1 0 3 4x 2x 0 2 9 x 3 7 x x 1