In a previous lesson, we solved certain quadratic equations by taking the square root of both sides of the equation. x = 36 (x 3) = 8 x = ± 36 x 3 = ± 8 x = ±6 x = 3 ± Taking the square root of both sides of an equation is the equivalent of raising both sides to the power of 1. So to eliminate an exponent of, we raised both sides of an equation to the power of 1. x = 9 (x ) 1 = ±(9) 1 x = ±3 The opposite is also be true; to eliminate an exponent of 1, we can raise both sides of an equation to the power of, or square both sides. x = 3 ( x) = (3) x = 9 This is what s known as the principle of powers. Principle of powers: - two equivalent expressions will still be equivalent if they are raised to the same power o x a = y a if and only if x = y - this is another example of the #1 rule of algebra; whatever you do to one side of an equation, you must do to the other
Steps for Solving Radical Equations: 1. isolate the radical. raise both sides of the equation to a power that will eliminate the radical (principle of powers) a. if your radical is a square root (like we ll see in these notes), square both sides b. if your radical is a cubed root (like we ll see in the next set of notes), cube both sides 3. solve a. after eliminating the radical, solve the equation that remains 4. check your answers by plugging them back into the original equation a. it is imperative that you check your solutions when solving radical equations with even roots, such as the square roots in this set of notes, as well as square roots, 4 th roots, and 6 th roots in the next set of notes Example 1: Solve the following equations for x and enter exact answers only (no decimal approximations). If there is more than one solution, separate your answers with commas. If there are no real solutions, enter NO SOLUTION. x + 5x + 19 = 1 5x + 19 = 1 x Similar to rational equations (equations with fractions), radical equations have a restricted domain. That means there is a limited set of values that can take the place of x, and the restriction is based the fact the radicals with even roots (such as square roots) cannot have a radicand that s negative and cannot be equal to a negative value. So in this case, the equation 5x + 19 = 1 x tells us that 5x + 19 0 AND 1 x 0, because 5x + 19 is the radicand of a square root and 1 x is equal to a square root.
Example 1 continued: x + 5x + 19 = 1 5x + 19 = 1 x ( 5x + 19 ) = ( 1 x) 5x + 19 = ( 1 x)( 1 x) 5x + 19 = 1 + x + x 0 = x 3x 18 0 = x + 3x 6x 18 0 = x(x + 3) 6(x + 3) 0 = (x + 3)(x 6) 0 = x + 3 ; 0 = x 6 3 = x ; 6 = x Now we have to check to verify that both answers make the original equation true.
x = 3 x = 6 x + 5x + 19 = 1 3 + 5( 3) + 19 = 1 3 + 15 + 19 = 1 3 + 4 = 1 3 + = 1 1 = 1 x + 5x + 19 = 1 6 + 5(6) + 19 = 1 6 + 30 + 19 = 1 6 + 49 = 1 6 + 7 = 1 13 1 Since x = 3 made the original equation true, 3 is a valid answer. Since x = 6 did not make the original equation true, 6 is not a valid answer. x = 3 Again, the reason we must check our answers is because equations with square roots, just like equations with fractions, have restrictions on what can replace x. In Example 1, 3 is a value that can replace x while 6 is not. And just like when we solve rational equations, you can find the restricted values first, before solving the equation, or you can solve the equation first and then plug your answers back in. Either way at some point, you must determine what values are acceptable to replace x. Finding the restricted values prior to solving Example 1 would have meant solving 5x + 19 0 AND 1 x 0, which results in the compound inequality 3.8 x 1. Of the answers we came up with, x = 3 is the only one that satisfies that inequality.
Example : Solve the following equations for x and enter exact answers only (no decimal approximations). If there is more than one solution, separate your answers with commas. If there are no real solutions, enter NO SOLUTION. 4 + x = x 4 + x = x x = x 4 ( x) = (x 4) x = (x 4)(x 4) x = 4x 16x + 16 0 = 4x 15x + 14 0 = 4x 7x 8x + 14 0 = x(4x 7) (4x 7) 0 = (4x 7)(x ) 0 = 4x 7 ; 0 = x 7 4 = x ; = x ALWAYS start by isolating the radical you are given in the equation. In this case I multiply both sides by to eliminate the fraction, then I subtract 4 from both sides. Once the radical is isolated, eliminate it by using an exponent that is equivalent to the index of the radical. In this example I have a square root, so I use a power of. Once the radical is eliminated, I can remove parentheses, combine like terms, and solve the resulting equation (in this case a quadratic equations). Keep in mind there are multiple ways to solve a quadratic equation, so if you don t want to factor like I did, you are welcome to use a different method. Now we have to check to verify that both answers make the original equation true.
x = 7 4 4 + x 4 + 7 4 4 + 1 4 = 7 4 4 + 1 = 7 4 9 = 7 4 9 4 = 7 4 = x = 7 4 x = 4 + x 4 + 4 + 0 4 + 0 = = 4 = = = x = Solving the quadratic equation from the previous page produces two answers, but because radicals with even roots have restricted domains, one of those answer might not be valid. So we must plug both answers back into the original equation to determine whether one, both, or neither answer is valid. Since x = made the original equation true, is a valid answer. Since x = 7 did not make the original equation true, 7 is not a valid answer. 4 4 x = Again, the reason we must check our answers is because equations with square roots, just like equations with fractions, have restrictions on what can replace x. In Example, is a value that can replace x while 7 is not. 4 And just like when we solve rational equations, you can find the restricted values first, before solving the equation, or you can solve the equation first and then plug your answers back in. Either way at some point, you must determine what values are acceptable to replace x.
Example 3: Solve the following equations for x and enter exact answers only (no decimal approximations). If there is more than one solution, separate your answers with commas. If there are no real solutions, enter NO SOLUTION. a. x = 4 + 4x 19 b. x 18x 81 = 0 b.
c. 7x 4 = 7 + 8x d. x + 9 = 4x 4 d.
Answers to Examples: 3a. x = 5, 7 ; 3b. x = 9 ; 3c. NO SOLUTION ; 3d. x = 7 ;