How Do We Solve Radical Equations? • Do Now: Simplify the given expression. 1. 2. Radical Equations An equation in which a variable occurs in the radicand is called a radical equation. It should be noted, that when solving a radical equation algebraically, extraneous roots may be introduced when both sides of an equation are squared. Therefore, you must check your solutions for a radical equation. Check: L.S. R.S. Solve: √ x - 3 - 3 = 0 x≥3 0 √x-3 -3 √x-3 =3 (√ x - 3 )2 = (3)2 √ 12 - 3 - 3 3-3 x-3=9 0 x = 12 Therefore, the solution is x = 12. If 2 x 1 2 5, then x is equal to (1) 1 (2) 2 (3) 5 (4) 4 2x 1 2 5 2x 1 3 2x 1 9 2 x 10 x5 What is the solution of the equation 2x 3 3 6 ? (1) 42 (2) 39 (3) 3 (4) 6 2x 3 3 6 2x 3 9 2 x 3 81 2 x 84 x 42 Solving Radical Equations 4 + √ 4 + x2 = x Check: √ 4 + x2 = x - 4 (√ 4 + x2)2 = (x - 4)2 4 + x2 = x2 - 8x + 16 8x = 12 3 x 2 3 13 , the solution of Since 2 2 3 x= is extraneous. Therefore, 2 there are no real roots. 4 4x 2 x 2 3 4 4 2 3 2 9 4 4 4 25 4 4 5 4 2 13 2 ≠ 3 2 The solution set of the equation x6 x is (1) {–2,3} (2) {–2} (3) {3} (4) { } x6 x x6 x 2 x x6 0 2 ( x 3)( x 2) 0 x 3 x 2 What is the solution set of the equation 9 x 10 x ? (1) {-1} (2) {9} (3) {10} (4) {10, -1} 9 x 10 x 9 x 10 x 2 x 9 x 10 0 2 ( x 10)( x 1) 0 x 10 x 1 x = -1 is an extraneous solution. Solving Radical Equations Solve 2x 4 x 7 0. 2x 4 x7 2x 4 x 7 2 2x 4 x 7 2(3) 4 3 7 10 10 0 Set up the equation so that there will be one radical on each side of the equal sign. 2 2x + 4 = x + 7 x=3 L.S. x ≥ -2 Square both sides. Simplify. R.S. 0 Verify your solution. Therefore, the solution is x = 3. Squaring a Binomial (a + 2)2 = a2 + 4a + 4 ( 5 + √x - 2 )2 5 x 2 2 10 x 2 5 x 2 2 25 10 x 2 (x 2) x 23 10 x 2 Note that the middle term is twice the product of the two terms of the binomial. The middle term will be twice the product of the two terms. A final concept that you should know: (a√x + b)2 = a2(x + b) = a2x + ab Solving Radical Equations Solve 5x 1 3x 5 2. 5x 1 2 3x 5 5x 1 2 2 3x 5 Square both sides of the equation. 5x 1 4 4 3x 5 (3x 5) Use Foil. 5x 1 3x 1 4 3x 5 Simplify. 2x 2 4 3x 5 Simplify by dividing by a common factor of 2. x 1 2 3x 5 x 1 2 3x 5 2 x 2x 1 4(3x 5) 2 2 Set up the equation so that there will be only one radical on each side of the equal sign. 2 Square both sides of the equation. Use Foil. Solving Radical Equations x 2x 1 4(3x 5) Distribute the 4. x 2x 1 12x 20 Simplify. 2 2 x 10x 21 0 (x 3)(x 7) 0 x - 3 = 0 or x - 7 = 0 x = 3 or x=7 2 L.S. 5x 1 3x 5 Factor the quadratic. Solve for x. Verify both solutions. R.S. L.S. 2 5x 1 3x 5 R.S. 5(3) 1 3(3) 5 5(7) 1 3(7) 5 42 64 2 2 2 One more to see another extraneous solution: Thealgebraically radical is already but DOESisolated NOT 3xa solution 1 that x you 3 find make a true statement when you substitute it back 2 2 Square both sides into the equation. You must square the whole side NOT each term. 3x 1 x 3 This must be FOILed 3 x 1 x 6 x 9 You MUST check answers 2 you have a quadratic these Since equation (has an x2 x 9 x 8 0 term) get everything on one side = 0 and see if 3 1 1 1 3 you can factor this 3 8 8 3 x 8x 1 0 Doesn't work! It checks! x 8, x 1 52 52 Extraneous 2 Let's try another one: 2 x 1 1 3 2 x 1 1 0 -1 First isolate the radical -1 1 3 3 1 3 2x 1 1 -1 -1 2 x 2 x 1 Remember that the 1/3 Now is same a 1/3thing powersince meansitthe as a cube root. power this means the same as a cube root so cube both sides Now solve for x Let's check this answer 21 1 1 0 0 0 It checks! 3 Graphing a Radical Function Graph y x 2. The domain is x > -2. The range is y > 0. Solving a Radical Equation Graphically Solve The solution will be the intersection of the graph x 3 3 0. y x33 and the graph of y = 0. The solution is x = 12. Check: L.S. x33 12 3 3 9 3 33 R.S. 0 Solving a Radical Equation Graphically Solve 5x 1 3x 5 2. The solution is x = 3 or x = 7. Solving Radical Inequalities Solve 7x 3 2 3. Note the radical 7x - 3 is defined only 3 when x . Find the values for which the graph of y 7x 3 2 7 is above the graph of y = 3. The graphs intersect at x = 4. x>4 Therefore, the solution is x > 4. Solving Radical Inequalities Solve x 1 3. x > -1 The graphs intersect at the point where x = 8. x ≥ -1 and x < 8 The solution is -1 < x and x < 8. This powerpoint was kindly donated to www.worldofteaching.com http://www.worldofteaching.com is home to over a thousand powerpoints submitted by teachers. This is a completely free site and requires no registration. Please visit and I hope it will help in your teaching.