9.4 Equations that Are Quadratic in Form

1 2001 McGraw-Hill Companies713 Equations that Are Quadraticin a radical equation that is Quadratic in a fourth degree equation that is Quadratic in formNOTE Let in (1).NOTE Let u x2in (2).NOTE Let u x2 xin (3).u 1xNOTE Notice that u2 xbecause u By intermediatesolutions we mean values for urather than for the originalvariable We square both sides ineach the following Equations :(1)(2)(x2 x)2 8(x2 x) 12 0(3)None of these Equations are Quadratic , yet each can be readily solved by using the following Quadratic Equations to the original three 5u 3 0(4)u2 4u 3 0(5)u2 8u 12 0(6)In each case, a simple substitution has been made that resulted in a Quadratic that can be rewritten in this manner are said to be Equations in Quadratic 4x2 3 02x 51x 3 1 Solving a Radical substituting ufor we have2u2 5u 3 0 Factoring yields(2u 3)(u 1) 0which gives the intermediate solutionsWe must now solve for xand then check our solutions.

2 Because we can writex 1To check these solutions, we again simply substitute these values into the original should verify that each is a valid 941x 32 or 1x 11x u,u 32 or u 11x,2x 51x 3 0714 CHAPTER9 QUADRATICEQUATIONS, FUNCTIONS, ANDINEQUALITIES 2001 McGraw-Hill CompaniesFor certain Equations in Quadratic form, we can either solve by substitution (as we havedone above) or solve directly by treating the equation as Quadratic in some other power ofthe variable (in the case of the equation of the following example, x2). In our next example,we show both methods of YOURSELF 1 Solve 3x 81x 4 2 Solving a Fourth Degree Equation(a)Solve x4 4x2 3 0 by substitution:Let u x2. Thenu2 4u 3 0 Factoring, we have(u 1)(u 3) 0sou 1oru 3 Given these intermediate solutions, because u x2, we can writex2 1orx2 3which, by using the square root method, yields the four solutionsx 1orx We can check each of these solutions by substituting into the original equation.

3 When wedo so, we find that all four are valid solutions to the original equation.(b)Solve x4 4x2 3 0 treating the equation as Quadratic in x2, we can factor the left member, to write(x2 1)(x2 3) 0 This gives us the two equationsx2 1 0orx2 3 0 Nowx2 1orx2 3 Again, we have the four possible solutionsx 1orx All check when they are substituted into the original Notice that u2 (x2)2 x4 NOTE There are foursolutionsto the original We apply the squareroot YOURSELF 2 Solve x4 9x2 20 0by substitution and by factoring THATAREQUADRATIC 2001 McGraw-Hill CompaniesIn the following example, a binomial is replaced with uto make it easier to proceed withthe Write in standard Factor the 3 Solving a Fourth Degree EquationSolve.(x2 x)2 8(x2 x) 12 0 Because of the repeated factor x2 x, we substitute ufor x2 x. Factoring the resultingequationu2 8u 12 0gives(u 6)(u 2) 0 Sou 6oru 2We now have two intermediate solutions to work with.

4 Because u x2 x, we have twocases:If u 6, thenIf u 2, thenx2 x 6x2 x 2x2 x 6 0x2 x 2 0(x 3)(x 2) 0(x 2)(x 1) 0x 3orx 2x 2orx 1 The Quadratic Equations now yield four solutions that we must check. Substituting into theoriginal equation, you will find that all four are valid YOURSELF 3 Solve for x:(x2 2x)2 11(x2 2x) 24 0To summarize our work with Equations in Quadratic form, two approaches are com-monly used. The first involves substitution of a new intermediate variable to make the orig-inal equation Quadratic . The second solves the original equation directly by treating theequation as Quadratic in some other power of the original variable. The following algo-rithms outline the two 1 Make an appropriate substitution so that the equation 2 Solve the resulting equation for the intermediate 3 Use the intermediate values found in step 2 to find possible solutionsfor the original 4 Check the solutions of step 3 by substitution into the original by Step:Solving by Substitution716 CHAPTER9 QUADRATICEQUATIONS, FUNCTIONS, ANDINEQUALITIES 2001 McGraw-Hill CompaniesStep 1 Treat the original equation as Quadratic in some power of the variable,and 2 Solve the resulting 3 Check the solutions of step 2 by substitution into the original by Step:Solving by FactoringCHECK YOURSELF 2, 3.

5 1, 2, 3, 4 15 49, 4 2001 McGraw-Hill CompaniesExercisesSolve each of the following Equations by factoring directly and then applying the zeroproduct 9x2 20 7t2 12 x2 12 7x2 18 9w2 4 5x2 2 4x2 4 6y2 9 16x2 12 9x2 5 4z2 70 10y2 8 20t2 81 0 Solve each of the following x2 12 w2 12 y2 15 5x2 9 20 64 6 8 Section Date 2001 McGraw-Hill 8 9 24 25 023.(x 2)2 3(x 2) 10 024.(w 1)2 5(w 1) 6 025.(x2 2x)2 3(x2 2x) 2 026.(x2 4x)2 (x2 4x) 12 0 Solve each of the following Equations by any 41 6 028.(x 1) 6 8 029.(w 3)2 2(w 3) 1530.(x2 4x)2 7(x2 4x) 12 5y2 29t2 25 0An equation involving rational exponents may sometimes be solved by substitution. Forinstance, to solve an equation of the form ax1/2 bx1/4 c 0, make the substitutionu x1/4. Note that u2 (x1/4)2 x2/4 x1 the suggestion above to solve each of the following Equations .

6 Be sure to check 4x1/4 3 5x1/4 6 x1/4 x1/4 1 2x1/3 3 x1/5 61x Equations involving rational expressions can also be solved by the method of sub-stitution. For instance, to solve an equation of the formmake the substitution Note that Use the suggestion above to solve the following each of the following sum of an integer and twice its square root is 24. What is theinteger? sum of an integer and 3 times its square root is 40. Find sum of the reciprocal of an integer and the square of itsreciprocal is . What is the integer? difference between the reciprocal of an integer and thesquare of its reciprocal is Find the (x 3)2 2x 3 11(x 2)2 1x 2 61(x 1)2 5x 1 4 03x2 5x 22x2 1x 31x2 6x 8 0u2 1x 2 bx c 0 2001 McGraw-Hill 2001 McGraw-Hill CompaniesAnswers1. 2 3. 2i i 13. 0, 15. 2, 16, 256 21. 81 23. 7, 0 25. 1, 1 i 27. 36 29. 0, 8 1, 81 35. 16 37. 27, 1 53, 52 3, 12 14, 12 2, i162 1102, i13 i23 151715, 163, i16 12 122, 2 13,15,720