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1.3 solving equations

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    1.3 solving equations 1.3 solving equations Presentation Transcript

    • Solving Equations
    • Solving Equations
      Two expressions set equal to each other is called
      an equation.
    • Solving Equations
      Two expressions set equal to each other is called
      an equation. To solve an equation means to find vealue(s) for the variables that makes the equation true.
    • Solving Equations
      Two expressions set equal to each other is called
      an equation. To solve an equation means to find vealue(s) for the variables that makes the equation true. Equations made from polynomial expressions, rational expressions, or algebraic expressions are called polynomial equations, rational equations, or algebraic equations respectively.
    • Solving Equations
      Two expressions set equal to each other is called
      an equation. To solve an equation means to find vealue(s) for the variables that makes the equation true. Equations made from polynomial expressions, rational expressions, or algebraic expressions are called polynomial equations, rational equations, or algebraic equations respectively.
      We solve polynomial equations by factoring.
    • Solving Equations
      Two expressions set equal to each other is called
      an equation. To solve an equation means to find vealue(s) for the variables that makes the equation true. Equations made from polynomial expressions, rational expressions, or algebraic expressions are called polynomial equations, rational equations, or algebraic equations respectively.
      We solve polynomial equations by factoring.
      To solve other algebraic equations, we have to transform them into polynomial equations.
    • Solving Equations
      Two expressions set equal to each other is called
      an equation. To solve an equation means to find vealue(s) for the variables that makes the equation true. Equations made from polynomial expressions, rational expressions, or algebraic expressions are called polynomial equations, rational equations, or algebraic equations respectively.
      We solve polynomial equations by factoring.
      To solve other algebraic equations, we have to transform them into polynomial equations.
      Example A: 3x2 – 2x = 8
    • Solving Equations
      Two expressions set equal to each other is called
      an equation. To solve an equation means to find vealue(s) for the variables that makes the equation true. Equations made from polynomial expressions, rational expressions, or algebraic expressions are called polynomial equations, rational equations, or algebraic equations respectively.
      We solve polynomial equations by factoring.
      To solve other algebraic equations, we have to transform them into polynomial equations.
      Example A: 3x2 – 2x = 8
      Set one side to 0, 3x2 – 2x – 8 = 0
    • Solving Equations
      Two expressions set equal to each other is called
      an equation. To solve an equation means to find vealue(s) for the variables that makes the equation true. Equations made from polynomial expressions, rational expressions, or algebraic expressions are called polynomial equations, rational equations, or algebraic equations respectively.
      We solve polynomial equations by factoring.
      To solve other algebraic equations, we have to transform them into polynomial equations.
      Example A: 3x2 – 2x = 8
      Set one side to 0, 3x2 – 2x – 8 = 0 factor this;
      (3x + 4)(x – 2) = 0
    • Solving Equations
      Two expressions set equal to each other is called
      an equation. To solve an equation means to find vealue(s) for the variables that makes the equation true. Equations made from polynomial expressions, rational expressions, or algebraic expressions are called polynomial equations, rational equations, or algebraic equations respectively.
      We solve polynomial equations by factoring.
      To solve other algebraic equations, we have to transform them into polynomial equations.
      Example A: 3x2 – 2x = 8
      Set one side to 0, 3x2 – 2x – 8 = 0 factor this;
      (3x + 4)(x – 2) = 0 extract answers;
      x = -4/3, 2
    • Quadratic Formula and Discriminant
      Quadratic Formula (QF):
    • Quadratic Formula and Discriminant
      Quadratic Formula (QF):
      The roots for the equation ax2 + bx + c = 0 are
      –b ± b2 – 4ac
      x =
      2a
    • Quadratic Formula and Discriminant
      Quadratic Formula (QF):
      The roots for the equation ax2 + bx + c = 0 are
      b2 – 4ac is the discriminant because its value indicates the type of roots we have.
      –b ± b2 – 4ac
      x =
      2a
    • Quadratic Formula and Discriminant
      Quadratic Formula (QF):
      The roots for the equation ax2 + bx + c = 0 are
      b2 – 4ac is the discriminant because its value indicates the type of roots we have.
      I. If b2 – 4ac > 0, we have real roots,
      –b ± b2 – 4ac
      x =
      2a
    • Quadratic Formula and Discriminant
      Quadratic Formula (QF):
      The roots for the equation ax2 + bx + c = 0 are
      b2 – 4ac is the discriminant because its value indicates the type of roots we have.
      I. If b2 – 4ac > 0, we have real roots, furthermore if
      b2 – 4ac is a perfect square, the roots are rational.
      –b ± b2 – 4ac
      x =
      2a
    • Quadratic Formula and Discriminant
      Quadratic Formula (QF):
      The roots for the equation ax2 + bx + c = 0 are
      b2 – 4ac is the discriminant because its value indicates the type of roots we have.
      I. If b2 – 4ac > 0, we have real roots, furthermore if
      b2 – 4ac is a perfect square, the roots are rational.
      II. If b2 – 4bc < 0, the roots are not real.
      –b ± b2 – 4ac
      x =
      2a
    • Quadratic Formula and Discriminant
      Quadratic Formula (QF):
      The roots for the equation ax2 + bx + c = 0 are
      b2 – 4ac is the discriminant because its value indicates the type of roots we have.
      I. If b2 – 4ac > 0, we have real roots, furthermore if
      b2 – 4ac is a perfect square, the roots are rational.
      II. If b2 – 4bc < 0, the roots are not real.
      –b ± b2 – 4ac
      x =
      2a
      Example B: Find the values k where the solutions are real for x2 + 2x + (2 – 3k) = 0
    • Quadratic Formula and Discriminant
      Quadratic Formula (QF):
      The roots for the equation ax2 + bx + c = 0 are
      b2 – 4ac is the discriminant because its value indicates the type of roots we have.
      I. If b2 – 4ac > 0, we have real roots, furthermore if
      b2 – 4ac is a perfect square, the roots are rational.
      II. If b2 – 4bc < 0, the roots are not real.
      –b ± b2 – 4ac
      x =
      2a
      Example B: Find the values k where the solutions are real for x2 + 2x + (2 – 3k) = 0
      We need b2 – 4ac > 0,
    • Quadratic Formula and Discriminant
      Quadratic Formula (QF):
      The roots for the equation ax2 + bx + c = 0 are
      b2 – 4ac is the discriminant because its value indicates the type of roots we have.
      I. If b2 – 4ac > 0, we have real roots, furthermore if
      b2 – 4ac is a perfect square, the roots are rational.
      II. If b2 – 4bc < 0, the roots are not real.
      –b ± b2 – 4ac
      x =
      2a
      Example B: Find the values k where the solutions are real for x2 + 2x + (2 – 3k) = 0
      We need b2 – 4ac > 0, i.e 4 – 4(2 – 3k) > 0.
    • Quadratic Formula and Discriminant
      Quadratic Formula (QF):
      The roots for the equation ax2 + bx + c = 0 are
      b2 – 4ac is the discriminant because its value indicates the type of roots we have.
      I. If b2 – 4ac > 0, we have real roots, furthermore if
      b2 – 4ac is a perfect square, the roots are rational.
      II. If b2 – 4bc < 0, the roots are not real.
      –b ± b2 – 4ac
      x =
      2a
      Example B: Find the values k where the solutions are real for x2 + 2x + (2 – 3k) = 0
      We need b2 – 4ac > 0, i.e 4 – 4(2 – 3k) > 0.
      – 4 + 12k > 0
      or k > 1/3
    • Equations of the Form xp/q = c
    • Equations of the Form xp/q = c
      Solve xp/q = c by raising both sides to the reciprocal exponent q/p and check your answers afterwards.
    • Equations of the Form xp/q = c
      Solve xp/q = c by raising both sides to the reciprocal exponent q/p and check your answers afterwards.
      Example C: Solve x-2/3 = 16
    • Equations of the Form xp/q = c
      Solve xp/q = c by raising both sides to the reciprocal exponent q/p and check your answers afterwards.
      Example C: Solve x-2/3 = 16
      Raise both sides to -3/2 power.
    • Equations of the Form xp/q = c
      Solve xp/q = c by raising both sides to the reciprocal exponent q/p and check your answers afterwards.
      Example C: Solve x-2/3 = 16
      Raise both sides to -3/2 power.
      x-2/3 = 16 
      (x-2/3)-3/2 = (16)-3/2
    • Equations of the Form xp/q = c
      Solve xp/q = c by raising both sides to the reciprocal exponent q/p and check your answers afterwards.
      Example C: Solve x-2/3 = 16
      Raise both sides to -3/2 power.
      x-2/3 = 16 
      (x-2/3)-3/2 = (16)-3/2
      x= 1/64 and it's a solution.
    • Equations of the Form xp/q = c
      Solve xp/q = c by raising both sides to the reciprocal exponent q/p and check your answers afterwards.
      Example C: Solve x-2/3 = 16
      Raise both sides to -3/2 power.
      x-2/3 = 16 
      (x-2/3)-3/2 = (16)-3/2
      x= 1/64 and it's a solution.
      Example D: Solve (2x – 3)3/2 = -8
    • Equations of the Form xp/q = c
      Solve xp/q = c by raising both sides to the reciprocal exponent q/p and check your answers afterwards.
      Example C: Solve x-2/3 = 16
      Raise both sides to -3/2 power.
      x-2/3 = 16 
      (x-2/3)-3/2 = (16)-3/2
      x= 1/64 and it's a solution.
      Example D: Solve (2x – 3)3/2 = -8
      Raise both sides to 2/3 power.
    • Equations of the Form xp/q = c
      Solve xp/q = c by raising both sides to the reciprocal exponent q/p and check your answers afterwards.
      Example C: Solve x-2/3 = 16
      Raise both sides to -3/2 power.
      x-2/3 = 16 
      (x-2/3)-3/2 = (16)-3/2
      x= 1/64 and it's a solution.
      Example D: Solve (2x – 3)3/2 = -8
      Raise both sides to 2/3 power.
      [(2x – 3)3/2]2/3 = (-8)2/3
      (2x – 3)3/2 = -8 
    • Equations of the Form xp/q = c
      Solve xp/q = c by raising both sides to the reciprocal exponent q/p and check your answers afterwards.
      Example C: Solve x-2/3 = 16
      Raise both sides to -3/2 power.
      x-2/3 = 16 
      (x-2/3)-3/2 = (16)-3/2
      x= 1/64 and it's a solution.
      Example D: Solve (2x – 3)3/2 = -8
      Raise both sides to 2/3 power.
      [(2x – 3)3/2]2/3 = (-8)2/3
      (2x – 3)3/2 = -8 
      (2x – 3) = 4
    • Equations of the Form xp/q = c
      Solve xp/q = c by raising both sides to the reciprocal exponent q/p and check your answers afterwards.
      Example C: Solve x-2/3 = 16
      Raise both sides to -3/2 power.
      x-2/3 = 16 
      (x-2/3)-3/2 = (16)-3/2
      x= 1/64 and it's a solution.
      Example D: Solve (2x – 3)3/2 = -8
      Raise both sides to 2/3 power.
      [(2x – 3)3/2]2/3 = (-8)2/3
      (2x – 3)3/2 = -8 
      (2x – 3) = 4
      2x = 7  x = 7/2
    • Equations of the Form xp/q = c
      Solve xp/q = c by raising both sides to the reciprocal exponent q/p and check your answers afterwards.
      Example C: Solve x-2/3 = 16
      Raise both sides to -3/2 power.
      x-2/3 = 16 
      (x-2/3)-3/2 = (16)-3/2
      x= 1/64 and it's a solution.
      Example D: Solve (2x – 3)3/2 = -8
      Raise both sides to 2/3 power.
      [(2x – 3)3/2]2/3 = (-8)2/3
      (2x – 3)3/2 = -8 
      (2x – 3) = 4
      2x = 7  x = 7/2
      Since x = 7/2 doesn't work because 43/2 = -8,
      there is no solution.
    • Radical Equations
    • Radical Equations
      Solve radical equations by squaring both sides to
      remove the square root.
    • Radical Equations
      Solve radical equations by squaring both sides to
      remove the square root. Do it again if necessary.
      Reminder: (A ± B)2 = A2± 2AB + B2
    • Radical Equations
      Solve radical equations by squaring both sides to
      remove the square root. Do it again if necessary.
      Reminder: (A ± B)2 = A2± 2AB + B2
      Example E: Solve
      x + 4 = 5x + 4
    • Radical Equations
      Solve radical equations by squaring both sides to
      remove the square root. Do it again if necessary.
      Reminder: (A ± B)2 = A2± 2AB + B2
      Example E: Solve
      x + 4 = 5x + 4 square both sides;
      (x + 4)2 = (5x + 4 )2
    • Radical Equations
      Solve radical equations by squaring both sides to
      remove the square root. Do it again if necessary.
      Reminder: (A ± B)2 = A2± 2AB + B2
      Example E: Solve
      x + 4 = 5x + 4 square both sides;
      (x + 4)2 = (5x + 4 )2
      x + 2 * 4 x + 16= 5x + 4
    • Radical Equations
      Solve radical equations by squaring both sides to
      remove the square root. Do it again if necessary.
      Reminder: (A ± B)2 = A2± 2AB + B2
      Example E: Solve
      x + 4 = 5x + 4 square both sides;
      (x + 4)2 = (5x + 4 )2
      x + 2 * 4 x + 16= 5x + 4 isolate the radical;
      8x = 4x – 12
    • Radical Equations
      Solve radical equations by squaring both sides to
      remove the square root. Do it again if necessary.
      Reminder: (A ± B)2 = A2± 2AB + B2
      Example E: Solve
      x + 4 = 5x + 4 square both sides;
      (x + 4)2 = (5x + 4 )2
      x + 2 * 4 x + 16= 5x + 4 isolate the radical;
      8x = 4x – 12 divide by 4;
      2x = x – 3
    • Radical Equations
      Solve radical equations by squaring both sides to
      remove the square root. Do it again if necessary.
      Reminder: (A ± B)2 = A2± 2AB + B2
      Example E: Solve
      x + 4 = 5x + 4 square both sides;
      (x + 4)2 = (5x + 4 )2
      x + 2 * 4 x + 16= 5x + 4 isolate the radical;
      8x = 4x – 12 divide by 4;
      2x = x – 3 square again;
      (2x)2 = (x – 3)2
    • Radical Equations
      Solve radical equations by squaring both sides to
      remove the square root. Do it again if necessary.
      Reminder: (A ± B)2 = A2± 2AB + B2
      Example E: Solve
      x + 4 = 5x + 4 square both sides;
      (x + 4)2 = (5x + 4 )2
      x + 2 * 4 x + 16= 5x + 4 isolate the radical;
      8x = 4x – 12 divide by 4;
      2x = x – 3 square again;
      (2x)2 = (x – 3)2
      4x = x2 – 6x + 9
    • Radical Equations
      Solve radical equations by squaring both sides to
      remove the square root. Do it again if necessary.
      Reminder: (A ± B)2 = A2± 2AB + B2
      Example E: Solve
      x + 4 = 5x + 4 square both sides;
      (x + 4)2 = (5x + 4 )2
      x + 2 * 4 x + 16= 5x + 4 isolate the radical;
      8x = 4x – 12 divide by 4;
      2x = x – 3 square again;
      (2x)2 = (x – 3)2
      4x = x2 – 6x + 9
      0 = x2 – 10x + 9
    • Radical Equations
      Solve radical equations by squaring both sides to
      remove the square root. Do it again if necessary.
      Reminder: (A ± B)2 = A2± 2AB + B2
      Example E: Solve
      x + 4 = 5x + 4 square both sides;
      (x + 4)2 = (5x + 4 )2
      x + 2 * 4 x + 16= 5x + 4 isolate the radical;
      8x = 4x – 12 divide by 4;
      2x = x – 3 square again;
      (2x)2 = (x – 3)2
      4x = x2 – 6x + 9
      0 = x2 – 10x + 9
      0 = (x – 9)(x – 1)
      x = 9, x = 1
    • Radical Equations
      Solve radical equations by squaring both sides to
      remove the square root. Do it again if necessary.
      Reminder: (A ± B)2 = A2± 2AB + B2
      Example E: Solve
      x + 4 = 5x + 4 square both sides;
      (x + 4)2 = (5x + 4 )2
      x + 2 * 4 x + 16= 5x + 4 isolate the radical;
      8x = 4x – 12 divide by 4;
      2x = x – 3 square again;
      (2x)2 = (x – 3)2
      4x = x2 – 6x + 9
      0 = x2 – 10x + 9
      0 = (x – 9)(x – 1)
      x = 9, x = 1 Only 9 is good.
    • Rational Equations
      Solve rational equations by clearing all denominators using the LCD.
    • Rational Equations
      Solve rational equations by clearing all denominators using the LCD. Check the answers afterwards.
    • Rational Equations
      Solve rational equations by clearing all denominators using the LCD. Check the answers afterwards.
      2
      4
      =
      + 1
      Example F: Solve
      x – 2
      x + 1
    • Rational Equations
      Solve rational equations by clearing all denominators using the LCD. Check the answers afterwards.
      2
      4
      =
      + 1
      Example F: Solve
      LCD = (x – 2)(x + 1), multiple the LCD to both sides of the equation:
      x – 2
      x + 1
    • Rational Equations
      Solve rational equations by clearing all denominators using the LCD. Check the answers afterwards.
      2
      4
      =
      + 1
      Example F: Solve
      LCD = (x – 2)(x + 1), multiple the LCD to both sides of the equation:
      (x – 2)(x + 1)* [ ]
      x – 2
      x + 1
      2
      4
      =
      + 1
      x + 1
      x – 2
    • Rational Equations
      Solve rational equations by clearing all denominators using the LCD. Check the answers afterwards.
      2
      4
      =
      + 1
      Example F: Solve
      LCD = (x – 2)(x + 1), multiple the LCD to both sides of the equation:
      (x – 2)(x + 1)* [ ]
      x – 2
      x + 1
      (x + 1)
      2
      4
      =
      + 1
      x + 1
      x – 2
    • Rational Equations
      Solve rational equations by clearing all denominators using the LCD. Check the answers afterwards.
      2
      4
      =
      + 1
      Example F: Solve
      LCD = (x – 2)(x + 1), multiple the LCD to both sides of the equation:
      (x – 2)(x + 1)* [ ]
      x – 2
      x + 1
      (x + 1)
      (x – 2)
      2
      4
      =
      + 1
      x + 1
      x – 2
    • Rational Equations
      Solve rational equations by clearing all denominators using the LCD. Check the answers afterwards.
      2
      4
      =
      + 1
      Example F: Solve
      LCD = (x – 2)(x + 1), multiple the LCD to both sides of the equation:
      (x – 2)(x + 1)* [ ]
      x – 2
      x + 1
      (x + 1)
      (x – 2)
      (x + 1)(x – 2)
      2
      4
      =
      + 1
      x + 1
      x – 2
    • Rational Equations
      Solve rational equations by clearing all denominators using the LCD. Check the answers afterwards.
      2
      4
      =
      + 1
      Example F: Solve
      LCD = (x – 2)(x + 1), multiple the LCD to both sides of the equation:
      (x – 2)(x + 1)* [ ]
      x – 2
      x + 1
      (x + 1)
      (x – 2)
      (x + 1)(x – 2)
      2
      4
      =
      + 1
      x + 1
      x – 2
      2(x + 1)= 4(x – 2)+ 1*(x + 1)(x – 2)
    • Rational Equations
      Solve rational equations by clearing all denominators using the LCD. Check the answers afterwards.
      2
      4
      =
      + 1
      Example F: Solve
      LCD = (x – 2)(x + 1), multiple the LCD to both sides of the equation:
      (x – 2)(x + 1)* [ ]
      x – 2
      x + 1
      (x + 1)
      (x – 2)
      (x + 1)(x – 2)
      2
      4
      =
      + 1
      x + 1
      x – 2
      2(x + 1)= 4(x – 2)+ 1*(x + 1)(x – 2)
      2x + 2 = 4x – 8 + x2 – x – 2
    • Rational Equations
      Solve rational equations by clearing all denominators using the LCD. Check the answers afterwards.
      2
      4
      =
      + 1
      Example F: Solve
      LCD = (x – 2)(x + 1), multiple the LCD to both sides of the equation:
      (x – 2)(x + 1)* [ ]
      x – 2
      x + 1
      (x + 1)
      (x – 2)
      (x + 1)(x – 2)
      2
      4
      =
      + 1
      x + 1
      x – 2
      2(x + 1)= 4(x – 2)+ 1*(x + 1)(x – 2)
      2x + 2 = 4x – 8 + x2 – x – 2
      2x + 2 = x2 + 3x – 10
    • Rational Equations
      Solve rational equations by clearing all denominators using the LCD. Check the answers afterwards.
      2
      4
      =
      + 1
      Example F: Solve
      LCD = (x – 2)(x + 1), multiple the LCD to both sides of the equation:
      (x – 2)(x + 1)* [ ]
      x – 2
      x + 1
      (x + 1)
      (x – 2)
      (x + 1)(x – 2)
      2
      4
      =
      + 1
      x + 1
      x – 2
      2(x + 1)= 4(x – 2)+ 1*(x + 1)(x – 2)
      2x + 2 = 4x – 8 + x2 – x – 2
      2x + 2 = x2 + 3x – 10
      0 = x2 + x – 12
    • Rational Equations
      Solve rational equations by clearing all denominators using the LCD. Check the answers afterwards.
      2
      4
      =
      + 1
      Example F: Solve
      LCD = (x – 2)(x + 1), multiple the LCD to both sides of the equation:
      (x – 2)(x + 1)* [ ]
      x – 2
      x + 1
      (x + 1)
      (x – 2)
      (x + 1)(x – 2)
      2
      4
      =
      + 1
      x + 1
      x – 2
      2(x + 1)= 4(x – 2)+ 1*(x + 1)(x – 2)
      2x + 2 = 4x – 8 + x2 – x – 2
      2x + 2 = x2 + 3x – 10
      0 = x2 + x – 12
      0 = (x + 4)(x – 3)  x = -4, 3
    • Rational Equations
      Solve rational equations by clearing all denominators using the LCD. Check the answers afterwards.
      2
      4
      =
      + 1
      Example F: Solve
      LCD = (x – 2)(x + 1), multiple the LCD to both sides of the equation:
      (x – 2)(x + 1)* [ ]
      x – 2
      x + 1
      (x + 1)
      (x – 2)
      (x + 1)(x – 2)
      2
      4
      =
      + 1
      x + 1
      x – 2
      2(x + 1)= 4(x – 2)+ 1*(x + 1)(x – 2)
      2x + 2 = 4x – 8 + x2 – x – 2
      2x + 2 = x2 + 3x – 10
      0 = x2 + x – 12
      0 = (x + 4)(x – 3)  x = -4, 3
      Both are good.
    • Factoring By Grouping
      Some polynomial may be factored by pulling out common factors twice. We call this factor by grouping.
    • Factoring By Grouping
      Some polynomial may be factored by pulling out common factors twice. We call this factor by grouping.
      Example G: Solve 2x3 – 3x2 – 8x + 12 = 0
    • Factoring By Grouping
      Some polynomial may be factored by pulling out common factors twice. We call this factor by grouping.
      Example G: Solve 2x3 – 3x2 – 8x + 12 = 0
      2x3 – 3x2 – 8x + 12 = 0
    • Factoring By Grouping
      Some polynomial may be factored by pulling out common factors twice. We call this factor by grouping.
      Example G: Solve 2x3 – 3x2 – 8x + 12 = 0
      2x3 – 3x2 – 8x + 12 = 0
      x2(2x – 3) – 4(2x – 3) = 0
    • Factoring By Grouping
      Some polynomial may be factored by pulling out common factors twice. We call this factor by grouping.
      Example G: Solve 2x3 – 3x2 – 8x + 12 = 0
      2x3 – 3x2 – 8x + 12 = 0
      x2(2x – 3) – 4(2x – 3) = 0
      (2x – 3)(x2 – 4) = 0
    • Factoring By Grouping
      Some polynomial may be factored by pulling out common factors twice. We call this factor by grouping.
      Example G: Solve 2x3 – 3x2 – 8x + 12 = 0
      2x3 – 3x2 – 8x + 12 = 0
      x2(2x – 3) – 4(2x – 3) = 0
      (2x – 3)(x2 – 4) = 0
      (2x – 3)(x – 2)(x + 2) = 0
    • Factoring By Grouping
      Some polynomial may be factored by pulling out common factors twice. We call this factor by grouping.
      Example G: Solve 2x3 – 3x2 – 8x + 12 = 0
      2x3 – 3x2 – 8x + 12 = 0
      x2(2x – 3) – 4(2x – 3) = 0
      (2x – 3)(x2 – 4) = 0
      (2x – 3)(x – 2)(x + 2) = 0
      So x = 2/3, 2, -2
    • Absolute Value Equations
      The absolute value of x is the distance measured from x to 0 and it is denoted as |x|.
    • Absolute Value Equations
      The absolute value of x is the distance measured from x to 0 and it is denoted as |x|. Because it is distance, |x| is nonnegative.
    • Absolute Value Equations
      The absolute value of x is the distance measured from x to 0 and it is denoted as |x|. Because it is distance, |x| is nonnegative.
      Algebraic definition of absolute value:
      {
      x if x is positive or 0.
      |x|=
      –x (opposite of x) if x is negative.
    • Absolute Value Equations
      The absolute value of x is the distance measured from x to 0 and it is denoted as |x|. Because it is distance, |x| is nonnegative.
      Algebraic definition of absolute value:
      {
      x if x is positive or 0.
      |x|=
      –x (opposite of x) if x is negative.
      Hence | -5 | = –(-5) = 5.
    • Absolute Value Equations
      The absolute value of x is the distance measured from x to 0 and it is denoted as |x|. Because it is distance, |x| is nonnegative.
      Algebraic definition of absolute value:
      {
      x if x is positive or 0.
      |x|=
      –x (opposite of x) if x is negative.
      Hence | -5 | = –(-5) = 5. Since the absolute value is never negative, an equation such as
      |x4 – 3x + 1 | = – 2 doesn't have any solution.
    • Absolute Value Equations
      The absolute value of x is the distance measured from x to 0 and it is denoted as |x|. Because it is distance, |x| is nonnegative.
      Algebraic definition of absolute value:
      {
      x if x is positive or 0.
      |x|=
      –x (opposite of x) if x is negative.
      Hence | -5 | = –(-5) = 5. Since the absolute value is never negative, an equation such as
      |x4 – 3x + 1 | = – 2 doesn't have any solution.
      Fact I. |x*y| = |x|*|y|. For example, |-2*3 | = |-2|*|3| = 6.
    • Absolute Value Equations
      The absolute value of x is the distance measured from x to 0 and it is denoted as |x|. Because it is distance, |x| is nonnegative.
      Algebraic definition of absolute value:
      {
      x if x is positive or 0.
      |x|=
      –x (opposite of x) if x is negative.
      Hence | -5 | = –(-5) = 5. Since the absolute value is never negative, an equation such as
      |x4 – 3x + 1 | = – 2 doesn't have any solution.
      Fact I. |x*y| = |x|*|y|. For example, |-2*3 | = |-2|*|3| = 6.
      Warning: In general |x ± y|  |x| ± |y|.
    • Absolute Value Equations
      The absolute value of x is the distance measured from x to 0 and it is denoted as |x|. Because it is distance, |x| is nonnegative.
      Algebraic definition of absolute value:
      {
      x if x is positive or 0.
      |x|=
      –x (opposite of x) if x is negative.
      Hence | -5 | = –(-5) = 5. Since the absolute value is never negative, an equation such as
      |x4 – 3x + 1 | = – 2 doesn't have any solution.
      Fact I. |x*y| = |x|*|y|. For example, |-2*3 | = |-2|*|3| = 6.
      Warning: In general |x ± y|  |x| ± |y|.
      For instance, | 2 – 3 |  |2| – |3|  |2| + |3|.
    • Absolute Value Equations
      Because |x±y|  |x|±|y|, we have to solve absolute value equations by rephrasing it into two equations without absolute value.
    • Absolute Value Equations
      Because |x±y|  |x|±|y|, we have to solve absolute value equations by rephrasing it into two equations without absolute value.
      Fact II: If |x| = a where a is a positive number, then x = a or x = –a.
    • Absolute Value Equations
      Because |x±y|  |x|±|y|, we have to solve absolute value equations by rephrasing it into two equations without absolute value.
      Fact II: If |x| = a where a is a positive number, then x = a or x = –a.
      Example H:
      a. If | x | = 3
    • Absolute Value Equations
      Because |x±y|  |x|±|y|, we have to solve absolute value equations by rephrasing it into two equations without absolute value.
      Fact II: If |x| = a where a is a positive number, then x = a or x = –a.
      Example H:
      a. If | x | = 3 then x = 3 or x= –3
    • Absolute Value Equations
      Because |x±y|  |x|±|y|, we have to solve absolute value equations by rephrasing it into two equations without absolute value.
      Fact II: If |x| = a where a is a positive number, then x = a or x = –a.
      Example H:
      a. If | x | = 3 then x = 3 or x= –3
      b. If | 2x – 3 | = 5 then
    • Absolute Value Equations
      Because |x±y|  |x|±|y|, we have to solve absolute value equations by rephrasing it into two equations without absolute value.
      Fact II: If |x| = a where a is a positive number, then
      x = a or x = –a. Also if |x| = |y| then x = y or x = –y.
      Example H:
      a. If | x | = 3 then x = 3 or x= –3
      b. If | 2x – 3 | = 5 then
      2x – 3 = 5 or 2x – 3 = –5
    • Absolute Value Equations
      Because |x±y|  |x|±|y|, we have to solve absolute value equations by rephrasing it into two equations without absolute value.
      Fact II: If |x| = a where a is a positive number, then
      x = a or x = –a. Also if |x| = |y| then x = y or x = –y.
      Example H:
      a. If | x | = 3 then x = 3 or x= –3
      b. If | 2x – 3 | = 5 then
      2x – 3 = 5 or 2x – 3 = –5
      c. Solve |2x – 3| = |3x + 1|
    • Absolute Value Equations
      Because |x±y|  |x|±|y|, we have to solve absolute value equations by rephrasing it into two equations without absolute value.
      Fact II: If |x| = a where a is a positive number, then
      x = a or x = –a. Also if |x| = |y| then x = y or x = –y.
      Example H:
      a. If | x | = 3 then x = 3 or x= –3
      b. If | 2x – 3 | = 5 then
      2x – 3 = 5 or 2x – 3 = –5
      c. Solve |2x – 3| = |3x + 1|
      Rewrite: 2x – 3 = 3x + 1 or 2x – 3 = –(3x + 1)
    • Absolute Value Equations
      Because |x±y|  |x|±|y|, we have to solve absolute value equations by rephrasing it into two equations without absolute value.
      Fact II: If |x| = a where a is a positive number, then
      x = a or x = –a. Also if |x| = |y| then x = y or x = –y.
      Example H:
      a. If | x | = 3 then x = 3 or x= –3
      b. If | 2x – 3 | = 5 then
      2x – 3 = 5 or 2x – 3 = –5
      c. Solve |2x – 3| = |3x + 1|
      Rewrite: 2x – 3 = 3x + 1 or 2x – 3 = –(3x + 1)
      4 = x or 2x – 3 = –3x – 1
    • Absolute Value Equations
      Because |x±y|  |x|±|y|, we have to solve absolute value equations by rephrasing it into two equations without absolute value.
      Fact II: If |x| = a where a is a positive number, then
      x = a or x = –a. Also if |x| = |y| then x = y or x = –y.
      Example H:
      a. If | x | = 3 then x = 3 or x= –3
      b. If | 2x – 3 | = 5 then
      2x – 3 = 5 or 2x – 3 = –5
      c. Solve |2x – 3| = |3x + 1|
      Rewrite: 2x – 3 = 3x + 1 or 2x – 3 = –(3x + 1)
      4 = x or 2x – 3 = –3x – 1
      4 = x or x = 2/5
    • Solving Equations