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