Solve Equations of the Quadratic Form - Tutorial




This is a tutorial on solving equations which may be reduced to quadratic form. Detailed solutions and explanations are included.

Review

A quadratic equation has the form

ax 2 + bx + c = 0

with a not equal to 0.

There are several methods to solve quadratic equations. In this tutorial we use the method of the quadratic formula and the method of factoring.


Example 1 : Find all real solutions to the equation.

x 4 + x 2 - 6 = 0

Solution to Example 1:

  • Given
    x 4 + x 2 - 6 = 0

  • Since (x 2) 2 = x 4, let u = x2 and rewrite the equation in term of u.
    u2 + u - 6 = 0

  • Factor the left side.
    (u + 3)(u - 2) = 0

  • Use the zero factor theorem to obtain simple equations.
    a) (u + 3) = 0
    b) u - 2 = 0

  • Solve equation a).
    u = -3

  • Solve equation b).
    u = 2

  • Use the fact that u = x 2, the first solution in u gives,
    x 2 = -3

  • and the second solution gives.
    x 2 = 2

  • The square of a real number cannot be negative and therefore the equation x 2 = -3 does not have any real solutions. The second equation is solved by extracting the square root and gives two solutions.
    x = sqrt(2)

    x = -sqrt(2)

Check Solutions

  1. x = sqrt(2)
    Left side of the equation = sqrt(2) 4 + sqrt(2) 2 - 6
    = 4 + 2 - 6
    = 0
    Right side of the equation = 0.

  2. x = -sqrt(2)
    Left side of the equation = (-sqrt(2)) 4 + (-sqrt(2)) 2 - 6
    = 4 + 2 - 6
    = 0
    Right side of the equation = 0.

Conclusion: The real solutions to the given equation are sqrt(2) and -sqrt(2)

Matched Exercise 1: Find all real solutions to the equation.

x 4 - 2 x 2 - 3 = 0

Answer


Example 2 : Find all real solutions to the equation

2 x + 3*sqrt(x) = 5

Solution to Example 2:

  • Given
    2 x + 3*sqrt(x) = 5

  • Note that sqrt(x) implies x has to be positive or zero. Since [ sqrt(x) ]2 = x , let u = sqrt(x) and rewrite the equation in term of u.
    2u2 + 3u = 5

  • Rewrite the equation with the right side equal to 0.
    2u2 + 3u - 5 = 0

  • Use the quadratic formula. The discriminant D is given by
    D = b2 - 4ac
    = (3)2 - 4(2)(-5)
    = 49

  • Use the quadratic formula to write the two solutions as follows.
    u1 = (-b + sqrt(D)) / 2a
    and
    u2 = (-b - sqrt(D)) / 2a

  • Substitute b, D and a by their values.
    u1 = (-3 + sqrt(49)) / 4
    and
    u2 = (-3 - sqrt(49)) / 4

  • Simplify the above expressions.
    u1 = 1     and     u2 = -5/2

  • We now use the fact that u = sqrt(x) and solve for x. The first solution u1 gives
    sqrt(x) = 1

  • Square both sides to obtain
    x = 1

  • The second solution u2 gives
    sqrt(x) = -5/2

  • This last equation have no real solutions since the square root of a real positive number is must be a real positive number.

Check Solutions x = 1 Left Side = 2 (1) + 3*sqrt(1)
= 5
Right Side = 5

Conclusion:
The real solution to the given equation is x = 1.

Matched Exercise 2. Find all real solutions to the equation.

x - 3*sqrt(x) - 4 = 0

Answer

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Updated: 2 April 2013

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