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

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.

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.

Answer

Example 2 : Find all real solutions to the equation

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.

Answer

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