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
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

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

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
u
2 - 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