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,
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.
x ^{4}  2 x ^{2}  3 = 0
Answer
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 u^{2} + 3 u = 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}  4 a c
= (3)^{2}  4(2)(5)
= 49

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

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

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

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

Square both sides to obtain
x = 1

The second solution u_{2} 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
More References and links
Solve Equations, Systems of Equations and Inequalities.