Equations reducible to quadratic form
Some equations that involve radical expressions can be solved by using the following
results.
The power principle
If P and Q are algebraic expressions and n is a positive integer, then every solution of P =
Q is a solution of Pn = Qn.
We will use the power principle to solve
x + 4 = 3.
Example 1:
x+4 =3
(
)
2
x + 4 = 32
x+4=9
x+4−4=9−4
x=5
Check:
We apply the power principle with n = 2 to remove the square root.
Simplify by raising both sides to power of 2
Subtract 4 from both sides to isolate variable term.
Simplify by combining like terms
x+4 =3
5 + 4 ?3
9 ?3
3=3
I’m using ? rather than = since I don’t know these are equal
5 checks
The solution is 5.
Some care must be taken when using the power principle, because the equation Pn = Qn
may have more solutions than the original P = Q. As an example, consider x = 3. the
only solution is the real number 3. Square each side of the equation to produce x2 = 9,
which has both 3 and –3 as solutions. The –3 is called an extraneous solution because it
is not a solution of the original equation x = 3. Extraneous solutions may be introduced
when we raise each side of an equation to an even power.
Let’s look at more examples of equations that can be reduced to quadratic form.
Example 2:
Solve x = 2 + 2 − x .
x=2+ 2−x
x−2=2+ 2−x −2
x−2= 2−x
( x − 2 )2 = (
)
2
Subtract 2 from both sides to isolate the radical
Now combine like terms to simplify
Now we will square both sides
2−x
Perform the operations
2
x − 4x + 4 = 2 − x
and simplify
2
x − 4x + 4 − 2 + x = 2 − x − 2 + x
x 2 − 3x + 2 = 0
(x − 2)(x − 1) = 0
x−2=0
x−2+2=0+2
x=2
Now we solve this equation. I will use factoring
If the product is zero, one or both factors is zero.
x −1= 0
x −1+1= 0 +1
x =1
Check:
x=2+ 2−x
x=2+ 2−x
2?2 + 2 − 2
1? 2 + 2 − 1
2?2 + 0
2=2
1? 2 + 1
1? 2 + 1
We see that 1 is not a solution because it does not make the original equation true.
Example 3
x + 1 − 2x − 5 = 1
This equation involves two radicals. If we simply square both sides as in our previous
examples, we still have radicals in our equation. We will see that by applying the power
principle more than once, we can remove the radicals one at a time.
Start by isolating one of the radicals. We will add
x + 1 − 2x − 5 = 1
(
2 x − 5 to both sides of the equation,
x + 1 − 2x − 5 + 2x − 5 = 1 + 2x − 5
x + 1 = 1 + 2x − 5
Now we will square both sides
) (
2
x + 1 = 1 + 2x − 5
)
2
The right side of our equation is a binomial to be squared.
Review your notes on squaring binomials.
x + 1 = 1 + 2 2 x − 5 + (2 x − 5) Now simplify
x + 1 = 2x − 4 + 2 2x − 5
We want to isolate the radical by adding –2x + 4 to both
sides
x + 1 − 2x + 4 = 2x − 4 + 2 2x − 5 − 2x + 4
− x + 5 = 2 2x − 5
The right side contains a radical, so we will square both
sides
(− x + 5)2 = (2
)
2
2x − 5
Square the binomial on the left side, square the right side
2
x − 10 x + 25 = 4(2 x − 5)
Remove parenthesis
2
x − 10 x + 25 = 8 x − 20
Add –8x + 20 to both sides
2
x − 10 x + 25 − 8 x + 20 = 8 x − 20 − 8 x + 20 Simplify
x 2 − 18 x + 45 = 0
I will now solve for x by factoring
(x − 3)(x − 15) = 0
If the product is zero, one or more of factors must be zero
x −3=0
x − 15 = 0
x −3+3=0+3
x − 15 + 15 = 0 + 15
x=3
x = 15
Check:
x + 1 − 2 x − 5 ?1
Proposed solutions are 3 and 15.
x + 1 − 2 x − 5 ?1
3 + 1 − 2 * 3 − 5 ?1
15 + 1 − 2 * 15 − 5 ? 1
4 − 6 − 1 ?1
16 − 30 − 5 ? 1
4 − 1 ?1
2 − 1?1
1=1
4 − 25 ? 1
4 − 5 ≠1
We find 3 is a solution but 15 is not a solution.
The equation 4x4 – 25x2 + 36 = 0 is said to be in quadratic form, which means it can be
written in the form au2 + bu + c = 0, a ≠ 0. This quadratic equation can be solved for u,
and then, using the relationship u = x2, we can find a solution for the original equation.
Example 4
Solve 4x4 – 25x2 + 36 = 0
Let u = x2, then u2 = x4 to produce the quadratic equation 4u2 – 25u + 36 = 0. We will
solve this equation using the techniques we’ve learned.
4u 2 − 25u + 36 = 0
(4u − 9)(u − 4) = 0
4u − 9 = 0
4u − 9 + 9 = 0 + 9
4u = 9
4u 9
=
4 4
9
u=
4
since our product is zero, one or more factors must be zero.
u−4=0
u−4+4=0+4
u=4
Substitute x2 for u to produce
9
x2 =
x2 = 4
4
9
x2 =
x2 = 4
4
3
x = ±2
x=±
2
Take the square root of both sides of each equation
Notice we have four solutions to check.
Check:
4 x 4 − 25 x 2 + 36 = 0
4 x 4 − 25 x 2 + 36 ? 0
4 * (− 2 ) − 25 * (− 2 ) + 36 ? 0
4 * 16 − 25 * 4 + 36 ? 0
64 − 100 + 36 ? 0
0=0
4
2
x * 2 4 − 25 * 2 2 + 36 ? 0
4 * 16 − 25 * 4 + 36 ? 0
64 − 100 + 36 ? 0
0=0
4 x 4 − 25 x 2 + 36 ? 0
4
2
 − 3
 − 3
4*
 − 25 * 
 + 36 ? 0
 2 
 2 
81
9
4 * − 25 * + 36 ? 0
16
4
81 225
−
+ 36 ? 0
4
4
− 144
+ 36 ? 0
4
− 36 + 36 ? 0
0=0
It is obvious that if we replace x with 3/2 the results will be equal. The solutions are –2, 3/2, 3/2, and 2.