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
ax ^{2} + bx + c = 0
with 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 = x^{2} and rewrite the equation in term of u.
u^{2} + 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

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.

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.
x ^{4}  2 x ^{2}  3 = 0
Answer
Example 2 : Find all real solutions to the
equation
2 x + 3*sqrt(x) = 5
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.
2u^{2} + 3u = 5

Rewrite the equation with the right side
equal to 0.
2u^{2} + 3u  5 = 0

Use the quadratic formula. The
discriminant D is given by
D = b^{2}  4ac
= (3)^{2}  4(2)(5)
= 49

Use the quadratic formula to write the two solutions as
follows.
u_{1} = (b + sqrt(D)) / 2a
and
u_{2} = (b  sqrt(D)) / 2a

Substitute b, D and a by their values.
u_{1} = (3 + sqrt(49)) / 4
and
u_{2} = (3  sqrt(49)) / 4

Simplify the above expressions.
u_{1} = 1 and u_{2} = 5/2

We now use the fact that u = sqrt(x) and solve for x. The first solution u_{1} gives
sqrt(x) = 1

Square both sides to obtain
x = 1

The second solution u_{2} 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.
x  3*sqrt(x)  4 = 0
Answer
More references and links on how to Solve Equations, Systems of Equations and Inequalities.
