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College Algebra: Class 4

College Algebra: Class 4. Radical Equations Objectives: Solve radical equations Solve equations quadratic in form Solve equations by factoring. Radical Equations and Equations in Quadratic Form. Radical Equations – equations that contain a variable under a radical

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College Algebra: Class 4

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  1. College Algebra: Class 4 • Radical Equations • Objectives: • Solve radical equations • Solve equations quadratic in form • Solve equations by factoring

  2. Radical Equations and Equations in Quadratic Form • Radical Equations – equations that contain a variable under a radical • Recall that fractional exponents are radicals. Ex: ½ power is a square root, 1/3 power is a cube root etc. • Beware of Extraneous Solutions – extra answers which do not make the equation true. • Must check all answers in the original equation.

  3. Process of Solving Radical Equations: • Isolate the radical • (x + 3)1/2 – 6 = 3 • (x + 3)1/2 = 9 • Raise both sides to a power to cancel the radical • ((x + 3)1/2)2 = 92 • X + 3 = 81 (square and square root cancel) • Solve remaining equation • X = 78 • Always check your answers. (78 + 3)1/2 – 6 = 3

  4. Examples: Solve each of the following. • (2x – 4)(1/3) – 2 = 0 • Isolate the radical: (2x – 4)1/3 = 2 • Raise both sides to the power to eliminate the radical: ((2x-4)1/3)3 = 23 • Radical and power cancel: 2x – 4 = 8 • Solve the remaining equation: 2x=12 so x=6 • Check: (12-4)1/3 – 2 = 0

  5. (x – 1)(1/2) = x – 7 • Radical is already isolated. • Square both sides. Remember to square the quantity • (x – 1) = (x – 7)2 • x– 1 = x2 – 14x + 49 • 0 = x2 – 15x + 50 • Solve the quadratic equation formed • 0 = (x – 10)(x – 5) • x – 10 = 0 x – 5 = 0 • x = 10 x = 5 • Check answers (10 – 1)1/2 = 10 – 7 • (5 – 1)1/2 = 5 – 7 5 does not check, solution is 10

  6. (2x + 3)(1/2) – (x + 2)(1/2) = 2 • (2x + 3)1/2 = 2 + (x + 2)1/2 (isolate one radical) • ((2x + 3)1/2)2 = (2 + (x + 2)1/2)2(square both sides) • 2x + 3 = 4 + 4(x + 2)1/2 + x + 2(distribute quantity) • 2x – x + 3 – 4 – 2 = 4(x + 2)1/2 (isolate 2nd radical) • X – 3 = 4(x + 2)1/2 (combine like terms) • (X – 3)2 = (4(x + 2)1/2)2 (square both sides) • X2 – 6x + 9 = 16(x + 2) (distribute quantity) • X2 – 22x – 23 = 0 (set equal to zero) • (x – 23)(x + 1) = 0 (solve quadratic equation) • X = 23 x = -1 -1 does not check, 23 is solution

  7. Process for Equations in Quadratic Form • Make sure the equation is in standard form. • Substitute a variable in for the quadratic value. Solve the quadratic equation formed by the method of your choice. • Replace the original value back in for the variable used in the second step. • Check

  8. Examples • (x + 2)2 + 11(x + 2) – 12 = 0 • U = x + 2 (set u = linear term) • U2 + 11u – 12 = 0 >>>> (u + 12)(u – 1)=0 (substitute in and solve) • U = -12 u = 1 • X + 2 = -12 x+ 2 = 1 (replace original variable back in and solve) • X = -14 x = -1 (check answers – both of these do check)

  9. X4 – 5x2 + 4 = 0 • u = x2(set u = linear term) • U2 – 5u + 4 = 0 (u – 4)(x – 1) = 0 (substitute in and solve) • U = 4 u = 1 • X2 = 4 x2 = 1 x = 2, -2, 1, -1 (replace original variable back in and solve)

  10. -7X3 – 8 = -X6 x6 – 7x3 – 8 = 0 u = x3 (set u = linear term) u2 – 7u – 8 = 0 (u – 8)(u + 1) = 0 (solve) u = 8 u = -1 x3 = 8 x3 = -1 (replace original variable back in) x = 2 x = -1 (both answers check)

  11. More examples • 1/(x + 1)2 = 1/(x + 1) + 2 • u = 1/(x + 1) • 1/(x + 1)2 – 1/(x + 1) – 2 = 0 Standard Form • u2 – u – 2 = 0 • Solution for u after substitution u = 2, u = -1 • X = -1/2 x = -2 • Both answers check

  12. Solve. If you have difficulties please see me for help. • X + 2x(1/2) – 3 = 0 • What would ‘u’ be substituted in for? (x1/2) • Solve then hit return to check your answer • X = 1 (9 did not check) • 2x-2 – 3x-1 - 4 = 0 • u = x-1 2u2 – 3u – 4 = 0 • Solve then hit return to check your answer • X = (-3 – sqrt 41) / 8 x = (-3 + sqrt 41) / 8

  13. Look over example 8 – Page 122. What would be another way to solve this equation? (POLY)Assn: Page 123 #7, 13, 19, 25, 33, 45, 49, 59, 95

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