Worked examples
The question
Question: solve x⁴ − 5x² + 4 = 0.
x4 − 5x2 + 4 = 0
1 Spot the pattern
x⁴ is x² times x², so x⁴ is x² squared.
x4 = (x2)2
So the equation has x² squared, and x², and a number.
It looks like a quadratic.
(x2)2 − 5(x2) + 4 = 0
2 Name u = x²
We give x² a short name: u.
Then x² squared is u squared.
u = x2
Now it is a quadratic in u: u² − 5u + 4 = 0.
We solve it like before.
u2 − 5u + 4 = 0
3 Factor
We need two numbers.
They multiply to 4, and they add to −5.
? × ? = 4
? + ? = −5
They multiply to a plus number, so the two numbers have the same sign.
They add to a minus number, so both are minus.
× is + → same signs: both −
We list the pairs that make 4, smallest first.
We add each pair.
We need 5.
1 and 4 give 5.
Yes!
With the signs: −1 and −4.
1 × 4: 1 + 4 = 5 ✓
So it is (u − 1)(u − 4) = 0.
(u − 1)(u − 4) = 0
4 One bracket is 0
Two numbers multiply to 0.
That only happens if one of them is 0.
So one bracket must be 0.
A × B = 0 → A = 0 or B = 0
We make each bracket 0.
Two small equations.
u − 1 = 0 or u − 4 = 0
In u − 1 = 0, we add 1 on both sides.
In u − 4 = 0, we add 4 on both sides.
u − 1 + 1 = 0 + 1 or u − 4 + 4 = 0 + 4
So u = 1 or u = 4.
u = 1 or u = 4
Careful!
We found u, not x!
u is x².
One more step to get x.
Careful: not x = 1 or x = 4 ✗
5 Back to x
u is x².
So x² = 1, or x² = 4.
x2 = 1 x2 = 4
x squared is a number.
So x is its square root, or minus it.
Both work, because minus times minus is plus.
( )² = k → ( ) = √k or −√k
1 times 1 is 1.
So x is plus or minus 1.
2 times 2 is 4.
So x is plus or minus 2.
= x = ±1 x = ±2
So the answers are −2, −1, 1 and 2.
= x = −2, −1, 1, 2
6 Check
Let's check: put x = 2 in.
−2 gives the same, because the powers are even.
x = ±2: 24 − 5(2)2 + 4 = 16 − 20 + 4 = 0
Now put x = 1 in.
−1 gives the same, because the powers are even.
x = ±1: 14 − 5(1)2 + 4 = 1 − 5 + 4 = 0
Every answer gives 0.
They all work!
Let's look back.
First: spot the pattern.
Then: name u = x².
Then: factor.
Then: one bracket is 0.
Then: back to x.
Last: check.
Answer = x = −2, −1, 1, 2
▶ Watch it solved
The question
Question: solve x⁶ − 9x³ + 8 = 0.
x6 − 9x3 + 8 = 0
1 Spot the pattern
x⁶ is x³ times x³, so x⁶ is x³ squared.
x6 = (x3)2
So the equation has x³ squared, and x³, and a number.
It looks like a quadratic.
(x3)2 − 9(x3) + 8 = 0
2 Name u = x³
We give x³ a short name: u.
Then x³ squared is u squared.
u = x3
Now it is a quadratic in u: u² − 9u + 8 = 0.
We solve it like before.
u2 − 9u + 8 = 0
3 Factor
We need two numbers.
They multiply to 8, and they add to −9.
? × ? = 8
? + ? = −9
They multiply to a plus number, so the two numbers have the same sign.
They add to a minus number, so both are minus.
× is + → same signs: both −
We list the pairs that make 8, smallest first.
We add each pair.
We need 9.
1 and 8 give 9.
Yes!
With the signs: −1 and −8.
1 × 8: 1 + 8 = 9 ✓
So it is (u − 1)(u − 8) = 0.
(u − 1)(u − 8) = 0
4 One bracket is 0
Two numbers multiply to 0.
That only happens if one of them is 0.
So one bracket must be 0.
A × B = 0 → A = 0 or B = 0
We make each bracket 0.
Two small equations.
u − 1 = 0 or u − 8 = 0
In u − 1 = 0, we add 1 on both sides.
In u − 8 = 0, we add 8 on both sides.
u − 1 + 1 = 0 + 1 or u − 8 + 8 = 0 + 8
So u = 1 or u = 8.
u = 1 or u = 8
Careful!
We found u, not x!
u is x³.
One more step to get x.
Careful: not x = 1 or x = 8 ✗
5 Back to x
u is x³.
So x³ = 1, or x³ = 8.
x3 = 1 x3 = 8
1 times 1 times 1 is 1.
A cube root has just one answer: x = 1.
2 times 2 times 2 is 8.
A cube root has just one answer: x = 2.
= x = 1 x = 2
So the answers are 1 and 2.
= x = 1, 2
6 Check
Let's check: put x = 1 in.
x = 1: 16 − 9(1)3 + 8 = 1 − 9 + 8 = 0
Now put x = 2 in.
x = 2: 26 − 9(2)3 + 8 = 64 − 72 + 8 = 0
Every answer gives 0.
They all work!
Let's look back.
First: spot the pattern.
Then: name u = x³.
Then: factor.
Then: one bracket is 0.
Then: back to x.
Last: check.
▶ Watch it solved
The question
Question: solve 4x⁴ − 5x² + 1 = 0.
4x4 − 5x2 + 1 = 0
1 Spot the pattern
x⁴ is x² times x², so x⁴ is x² squared.
x4 = (x2)2
So the equation has x² squared, and x², and a number.
It looks like a quadratic.
4(x2)2 − 5(x2) + 1 = 0
2 Name u = x²
We give x² a short name: u.
Then x² squared is u squared.
u = x2
Now it is a quadratic in u: 4u² − 5u + 1 = 0.
We solve it like before.
4u2 − 5u + 1 = 0
3 Find two numbers
There is a number in front of u², but it can still factor.
The number in front of u² is 4.
So we multiply 4 by 1: that is 4.
We need two numbers that multiply to 4 and add to −5.
4 × 1 = 4
? × ? = 4, ? + ? = −5
They multiply to a plus number, so the two numbers have the same sign.
They add to a minus number, so both are minus.
× is + → same signs: both −
We list the pairs that make 4, smallest first.
We add each pair.
We need 5.
1 and 4 give 5.
Yes!
With the signs: −1 and −4.
1 × 4: 1 + 4 = 5 ✓
4 Split the middle
We write −5u as −u − 4u.
It is the same amount.
4u2 − u − 4u + 1 = 0
5 Group in pairs
We put the first two together, and the last two together.
(4u2 − u) + (−4u + 1) = 0
From the first pair we take out u.
From the second pair we take out −1.
Both leave the same bracket.
(4u2 − u) = u(4u − 1)
(−4u + 1) = −1(4u − 1)
Now both parts have 4u − 1 in them.
u(4u − 1) − 1(4u − 1) = 0
6 Same bracket out
4u − 1 is in both parts, so we take it out once.
What is left, u − 1, goes in the other bracket.
ab + cb = b(a + c)
So we get (4u − 1)(u − 1).
(4u − 1)(u − 1) = 0
7 One bracket is 0
Two numbers multiply to 0.
That only happens if one of them is 0.
So one bracket must be 0.
A × B = 0 → A = 0 or B = 0
We make each bracket 0.
Two small equations.
4u − 1 = 0 or u − 1 = 0
In 4u − 1 = 0, we add 1 on both sides.
In u − 1 = 0, we add 1 on both sides.
4u − 1 + 1 = 0 + 1 or u − 1 + 1 = 0 + 1
Now we have 4u = 1 or u = 1.
4u = 1 or u = 1
To remove times 4, we divide both sides by 4.
So u = 1/4 or u = 1.
u = 14 or u = 1
Careful!
We found u, not x!
u is x².
One more step to get x.
Careful: not x = 1 over 4 or x = 1 ✗
8 Back to x
u is x².
So x² = 1/4, or x² = 1.
x2 = 14 x2 = 1
x squared is a number.
So x is its square root, or minus it.
Both work, because minus times minus is plus.
( )² = k → ( ) = √k or −√k
1/2 times 1/2 is 1/4.
So x is plus or minus 1/2.
1 times 1 is 1.
So x is plus or minus 1.
= x = ±12 x = ±1
So the answers are −1, −1/2, 1/2 and 1.
= x = −1, −12, 12, 1
9 Check
Let's check: put x = 1 in.
−1 gives the same, because the powers are even.
x = ±1: 4(1)4 − 5(1)2 + 1 = 4 − 5 + 1 = 0
Now put x = 1/2 in.
−1/2 gives the same, because the powers are even.
x = ±1 over 2: 4(12)4 − 5(12)2 + 1
= 14 − 54 + 1 = 0
Every answer gives 0.
They all work!
Let's look back.
First: spot the pattern.
Then: name u = x².
Then: find two numbers.
Then: split the middle.
Then: group in pairs.
Then: same bracket out.
Then: one bracket is 0.
Then: back to x.
Last: check.
Answer = x = −1, −12, 12, 1
▶ Watch it solved
