MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer Chabot Mathematics §7.6 2Var Radical Eqns

MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt 2 Bruce Mayer, PE Chabot College Mathematics Review §  Any QUESTIONS About §7.6 → Radical Equations  Any QUESTIONS About HomeWork §7.6 → HW MTH 55

MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt 3 Bruce Mayer, PE Chabot College Mathematics Radical Equations  A Radical Equation is an equation in which at least one variable appears in a radicand.  Some Examples:

MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt 4 Bruce Mayer, PE Chabot College Mathematics Solve Eqns with 2+ Rad. Terms 1.Isolate one of the radical terms. 2.Use the Exponent Power Rule 3.If a radical remains, perform steps (1) and (2) again. 4.Solve the resulting equation. 5.Check the possible solutions in the original equation.

MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt 5 Bruce Mayer, PE Chabot College Mathematics Example  Solve  SOLUTION

MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt 6 Bruce Mayer, PE Chabot College Mathematics Example  Solve  SOLUTION  Check 6 by Inspection → 3−2=1   Thus The number 6 checks and it IS the solution

MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt 7 Bruce Mayer, PE Chabot College Mathematics Example  Solve  SOLN One radical is isolated. We square both sides. Square both sides. Factoring Using the principle of zero products

MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt 8 Bruce Mayer, PE Chabot College Mathematics Example  Solve  Check: x = 3 x = 11   The numbers 3 and 11 check and are then confirmed as solutions.

MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt 9 Bruce Mayer, PE Chabot College Mathematics Example  Solve x – x + 6 = 5. Start by isolating one radical on one side of the equation by subtracting from each side. Then square both sides. x – x + 6 = 5 – x + 6 x + 7= 5 – – x x + 7= 5 – – x = 25 – 10 – x (– x + 6) x + 7 Twice the product of 5 and. – x + 6

MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt 10 Bruce Mayer, PE Chabot College Mathematics Example  Solve x – x + 6 = 5. This equation still contains a radical, so square both sides again. Before doing this, isolate the radical term on the right. = 25 – 10 – x (– x + 6) x + 7= 31 – x – 10 – x + 6 x + 7= –10 – x + 62 x – 24Subtract 31 and add x.= –5 – x + 6 x – 12Divide by 2. = –5 – x + 6( x – 12) 2 Square both sides again. 2

MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt 11 Bruce Mayer, PE Chabot College Mathematics Example  Solve x – x + 6 = 5. = –5 – x + 6( x – 12) 2 Square both sides again. 2 This equation still contains a radical, so square both sides again. = (–5) 2 – x + 6 x 2 – 24 x + 144( ab ) 2 = a 2 b 2 2 = 25 ( – x + 6 ) x 2 – 24 x = –25 x x 2 – 24 x + 144Distributive property = 0 x 2 + x – 6Standard form = 0( x + 3)( x – 2)Factor.

MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt 12 Bruce Mayer, PE Chabot College Mathematics Example  Solve x – x + 6 = 5. = 0( x + 3)( x – 2) x + 3 = 0 or x – 2 = 0Zero-factor property x = –3 or x = 2 Now finish solving the equation. Finally CHECK for Extraneous Solutions

MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt 13 Bruce Mayer, PE Chabot College Mathematics Example  Solve x – x + 6 = 5. Check each potential solution, –3 and 2, in the original equation. – –(–3) + 6 = 5? If x = –3, thenIf x = 2, then –(2) + 6 = 5? x – x + 6 = 5? = 5?9 + 4 = 5? 5 = 5 The solution set is { −3, 2 }. 

MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt 14 Bruce Mayer, PE Chabot College Mathematics The Principle of Square Roots  Recall the definition of the PRINCIPAL Square Root  For any NONnegative real number n, If x 2 = n, then

MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt 15 Bruce Mayer, PE Chabot College Mathematics Recall The Pythagorean Theorem  In any right triangle, if a and b are the lengths of the legs and c is the length of the hypotenuse, then a 2 + b 2 = c 2 a Leg Hypotenuse cb

MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt 16 Bruce Mayer, PE Chabot College Mathematics Example  Pythagorus  How long is a guy wire if it reaches from the top of a 14 ft pole to a point on the ground 8 ft from the pole?  SOLUTION 14 8 d We now use the principle of square roots. Since d represents a length, it follows that d is the positive square root of 260: Diagram

MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt 17 Bruce Mayer, PE Chabot College Mathematics Isosceles Right Triangle  When both legs of a right triangle are the same size, we call the triangle an isosceles right triangle. If one leg of an isosceles right triangle has length a then c a a

MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt 18 Bruce Mayer, PE Chabot College Mathematics Lengths for Isosceles Rt Triangles  The length of the hypotenuse in an isosceles right triangle is the length of a leg times 45 o a a

MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt 19 Bruce Mayer, PE Chabot College Mathematics Example  Isosceles Rt. Tri.  The hypotenuse of an isosceles right triangle is 8 ft long. Find the length of a leg. Give an exact answer and an approximation to three decimal places.  SOLUTION 45 o a a 8 8 = Exact answer: Approximation: ( after Rationalizing Divisor).

MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt 20 Bruce Mayer, PE Chabot College Mathematics 30°-60°-90° Triangle  A second special triangle is known as a 30°-60°-90° right triangle, so named because of the measures of its angles

MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt 21 Bruce Mayer, PE Chabot College Mathematics Lengths for 30/60/90 Rt Triangles  The length of the longer leg in a 30/60/90 right triangle is the length of the shorter leg times The hypotenuse is twice as long as the shorter leg. a 2a2a 60 o 30 o

MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt 22 Bruce Mayer, PE Chabot College Mathematics Example  30°-60°-90° Triangle  The shorter leg of a 30/60/90 right triangle measures 12 in. Find the lengths of the other sides. Give exact answers and, where appropriate, an approximation to three decimal places.  SOLUTION 12 2a2a 60 o 30 o The hypotenuse is twice as long as the shorter leg, so we have c = 2a= 2(12) = 24 in.

MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt 23 Bruce Mayer, PE Chabot College Mathematics Example  30°-60°-90° Triangle  SOLUTION  The length of the longer leg is the length of the shorter leg times This yields 12 2a2a 60 o 30 o = 12 Exact answer: Approximation:

MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt 24 Bruce Mayer, PE Chabot College Mathematics WhiteBoard Work  Problems From §7.6 Exercise Set 24, 34, 38, 48, 60  Astronomical Unit = Sun↔Earth Distance = km

MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt 25 Bruce Mayer, PE Chabot College Mathematics All Done for Today The Solar Star System

MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt 26 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer Chabot Mathematics Appendix –

MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt 27 Bruce Mayer, PE Chabot College Mathematics Graph y = |x|  Make T-table

MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt 28 Bruce Mayer, PE Chabot College Mathematics