Solve Equations That Can be Reduced
to the Quadratic Form - Tutorial

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

Review

A quadratic equation has the form
a x 2 + b x + c = 0
with the coefficient 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,
    u = x 2 = - 3
  • and the second solution gives.
    u = 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 = √2
    x = - √2

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

Conclusion: The real solutions to the given equation are √(2) and -√(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 √x = 5

Solution to Example 2:

  • Given
    2 x + 3 √x = 5
  • Note that √x implies x has to be positive or zero. Since [ √x ]2 = x , let u = √x and rewrite the equation in term of u.
    2 u2 + 3 u = 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 - 4 a c
    = (3)2 - 4(2)(-5)
    = 49
  • Use the quadratic formula to write the two solutions as follows.
    u1 = (- b + √D) / 2a
    and
    u2 = (-b - √D) / 2a
  • Substitute b, D and a by their values.
    u1 = (-3 + √(49)) / 4
    and
    u2 = (-3 - √(49)) / 4
  • Simplify the above expressions.
    u1 = 1    and    u2 = -5/2
  • We now use the fact that u = √x and solve for x. The first solution u1 gives
    √x = 1
  • Square both sides to obtain
    x = 1
  • The second solution u2 gives
    √x = - 5/2
  • This last equation have no real solutions since the square root of a real positive number is a real positive number.
Check Solutions x = 1 Left Side = 2 (1) + 3*√(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 √x - 4 = 0
Answer

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Solve Equations, Systems of Equations and Inequalities.






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