Solutions to Quadratic Equations With Rational Expressions

The solutions to questions on quadratic and rational equations are presented.

Solution to Question 1

  • The given equation
    x2 + 2x = -1
  • Write the above quadratic equation with right side equal to 0.
    x2 + 2x + 1 = 0
  • Factor the equation.
    (x + 1)2 = 0
  • Solve the equation to obtain the repeated solution.
    x = -1

Solution to Question 2

  • The given equation is a quadratic one
    x2 + 2 = x + 5
  • Write the equation with right side equal to 0.
    x2 - x - 3 = 0
  • Find the discriminant D of the quadratic equation.
    D = b 2 - 4 ac = 13
  • Solve the above equation to obtain 2 real solutions.
    x1 = [ 1 + √(13) ] / 2 and x2 = [ 1 - √(13) ] / 2

Solution to Question 3

  • The given equation
    -(x + 2)(x - 1)=3
  • Expand the left hand term of the equation.
    -x2 - x + 2 = 3
  • Rewrite the above equation with right side equal to 0.
    -x2 - x - 1 = 0
  • Find discriminant D.
    D = 1 - 4(-1)(-1) = -3
  • Since we are asked to find real solutions and the discriminant is negative, this equation has no real solutions.

Solution to Question 4

  • The equation to solve is
    (2x + 1) / (x + 2) = x - 1
  • The domain of the rational expression on the left of the equal sign of the equation is all real numbers except -2. Multiply both sides of the equation by (x + 2) and simplify.
    (2x + 1) = (x - 1)(x + 2)
  • Expand the right side, group like terms and write the equation in standard form.
    x2 - x - 3 = 0
  • Find discriminant D.
    D = 1 - 4(1)(-3) = 13
  • and the solutions are.
    x1 = [ 1 + √(13) ] / 2 and x2 = [ 1 - √(13) ] / 2

Solution to Question 5

  • The equation to solve is
    2 / (x + 1) - 1 / (x - 2) = -1
  • The LCM of the denominators of the rational expressions is.
    lcm = (x + 1)(x - 2)
  • We now multiply both sides of the equations by the lcm and simplify.
    2(x - 2) - 1(x + 1) = -1(x + 1)(x - 2)
  • Expand the right side and group.
    x - 5 = -x2 + x + 2
  • Write with right side equal to 0.
    x2 = 7
  • The solutions are.
    x1 = √7 and x2 = -√7

Solution to Question 6

  • The given equation is
    2(x - 2)2 - 6 = -2
  • Add 6 to both sides and simplify.
    2(x - 2)2 = 4
  • Divide both side by 2.
    (x - 2)2 = 2
  • Extract the square root to obtain.
    x - 2 = √2 and x - 2 = -√2
  • Solve for x both equations.
    x1 = 2 + √2 and x2 = 2 - √2

Solution to Question 7

  • The given equation is
    x / (x + 4) = -3 / (x - 2) + 18 / (x - 2) (x + 4)
  • The lcm of the denominators of the rationa expression is equal to
    lcm = (x + 4)(x - 2)
  • Multiply all terms by the lcm and simplify.
    x (x - 2) = -3 (x + 4) + 18
  • Expand and group.
    x 2 - 2x = -3x -12 + 18
  • Write the equation with right side equal to 0.
    x 2 + x - 6 = 0
  • Factor and solve.
    (x + 3)(x - 2) = 0
  • The solutions to the last equation are
    x = -3 and x = 2.
  • The solution x = 2 cannot be a solution to the given equation as it makes the denominator equal to 0. So the only solution to the given equation is x = -3.

Solution to Question 8

  • The equation to solve is
    x2 - 3(x - 3)2 = 2
  • Expand the term 3(x - 3)2 and group
    x2 - 3(x2 -6x + 9) = 2
    -2x2 + 18x - 27 = 2
  • Write the equation with right side equal to 0.
    -2x2 + 18x - 29 = 0
  • The discriminant D of the above quadratic equation is equal to .
    D = 92
  • The solutions to the given equation are
    x = [ 9 - √(23) ] / 2 and x = [ 9 + √(23) ]

Solution to Question 9

  • The equation to solve is
    1 / (x - 4) + 1 / (x + 4)= x2 / (x2 - 16)
  • The lcm of the denominators of the rational expressions is given by
    lcm = (x - 4)(x + 4) = x2 - 16
  • Multiply all terms by the lcm and simplify.
    x + 4 + x - 4 = x2
  • Simplify and write equation with right term equal to 0.
    x2 - 2 x = 0
  • Factor and solve
    x(x - 2) = 0
    x = 0 and x = 2 are the solutions.

Solution to Question 10

  • The equation to solve is
    -x / (x + 3) - x / (x - 3) = - 4 / (x2 - 9) - 1 / (x + 3)
  • The lcm of the denominators of the rational expressions is given by
    lcm = (x - 3)(x + 3) = x2 - 9
  • Multiply all terms of the equation by the lcm and simplify.
    -x(x - 3) - x(x + 3) = -4 -(x - 3)
  • Expand, group like terms and with right term equal to 0.
    2x2 - x - 1 = 0
  • Solve the above quadratic equation to obtain the solutions
    x = 1 and x = -1/2

More references and links to tutorials and questions on equations

Solve Quadratic Equations Using Discriminants
Tutorial on Equations of the Quadratic Form.
Math Problems, Questions and Online Self Tests.

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