Solve 4x^4 - 5x^2 + 1 = 0 Step by Step | Equations in quadratic form

Step by step

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