Solve Equations of the Quadratic Form

This is a tutorial with on solving equations which may be written in 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 Discriminants
and the method of factoring

Examples with Solutions

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 to Matched Exercise

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 to Matched Exercise

Solutions to Matched Exercises

Matched Exercise 1

Find all real solutions to the equation.
x 4 - 2 x 2 - 3 = 0

Let u = x 2
The above equation may be written as
u 2 - 2 u - 3 = 0
Solve the above for u to obtain the solutions
u = - 1 and u = 3
We now solve for x.
u = x 2 = - 1 , this equation has no real solutions.
u = x 2 = 3 gives x = √3 and x = - √3
The given equation has 2 real solutions.
x3 = √3
x4 = - √3

Matched Exercise 2

Find all real solutions to the equation.
x - 3 √x - 4 = 0
Let u = √x which gives u2 = x
The given equation is now written in terms of u as follows
u2 - 3 u - 4 = 0
Solve the above quadratic equation for u to obtain
u = - 1 and u = 4
Solve for x
u = √x = - 1 , this equation has no real solution √x is positive
u = √x = 4 , square both sides to obtain the solution
x = 16

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