At what times t is the particle at rest?

Step by step: way 1, factoring

The question

Question: solve t² − 5t + 4 = 0.

1 Factor

Factoring goes backwards.

We need two numbers.

They multiply to 4, and they add to −5.

Look at the sign in front of 4: it is plus.

So the two numbers have the same sign.

Now the sign in front of 5t: it is minus.

So both are minus.

We list the pairs that make 4, smallest first.

For each pair, we check two things: they multiply to 4, and they add to 5.

1 × 4 = 4.

1 + 4 = 5.

Both work!

So the two numbers are 1 and 4.

With the signs: −1 and −4.

So it is (t − 1)(t − 4) = 0.

2 One bracket is 0

Two numbers multiply to 0.

That only happens if one of them is 0.

So one bracket must be 0.

We make each bracket 0.

Two small equations.

In t − 1 = 0, we add 1 on both sides.

In t − 4 = 0, we add 4 on both sides.

So t = 1 or t = 4.

3 Check

Let's check: put t = 1 into the left side.

We get 0, the same as the right side.

It works!

Now put t = 4 into the left side.

We get 0 again, the same as the right side.

So both answers are right!

Let's look back.

First: factor.

Then: one bracket is 0.

Last: check.

Step by step: way 2, the quadratic formula

The question

Question: solve t² − 5t + 4 = 0.

1 Find a, b and c

So we use the formula.

It works for every equation ax² + bx + c = 0, with 0 on one side.

First we read the numbers.

a = 1 is with t².

b = −5 is with t.

c = 4 is the plain number.

2 Use the formula

Here is the formula.

People worked it out long ago, and the lesson on completing the square shows where it comes from.

± means we do it twice: once with plus, once with minus.

We put the numbers in, each in brackets.

−5 squared is 25.

4 times 1 times 4 is 16.

25 take away 16 is 9.

So under the root we have 9.

√9 is 3.

The plus or minus gives two answers.

So t = 4 or t = 1.

3 Check

Let's check: put t = 4 into the left side.

We get 0, the same as the right side.

It works!

Now put t = 1 into the left side.

We get 0 again, the same as the right side.

So both answers are right!

Let's look back.

First: find a, b and c.

Then: use the formula.

Last: check.

Step by step: way 3, completing the square

The question

Question: solve t² − 5t + 4 = 0 by completing the square.

Is t alone now?

Not yet: + 4 is still with t.

How do we remove it?

1 Move the 4

We want only the t parts on the left.

The + 4 is in the way.

So we take away 4 on both sides.

+ 4 and − 4 cancel each other.

They make zero.

0 minus 4 is −4.

Now we have t² − 5t = −4.

So after step 1: we took away 4, and now we have t² − 5t = −4.

2 Complete the square

We want the left side to be one bracket, squared.

A picture shows what is missing.

t² is a square: t by t.

We cut −5t in half: two equal parts, one beside the square and one below it.

Half of −5 is −5/2.

5 is odd, so the half is a fraction.

One corner is missing: −5/2 times −5/2.

Top times top, bottom times bottom: 25/4.

Minus times minus is plus.

So we add 25/4 to fill the corner.

We do it on both sides, so they stay equal.

3 Write it as a square

The full square is t − 5/2 on each side.

So the left side is (t − 5/2)².

To add, both need the same bottom.

The other bottom is 4.

−4 is −4/1.

Times 4 on top and bottom: −16/4.

Same bottom, so we add the tops: −16 plus 25 is 9.

So it is 9/4.

Careful!

Many people add 25/4 only on the left.

Then the two sides are not equal any more.

Now we have (t − 5/2)² = 9/4.

4 Square root of both sides

We want t alone.

But here we have (t − 5/2)².

(t − 5/2)² means (t − 5/2) times (t − 5/2).

So we need to remove the ².

To remove a square, we take the square root.

The square root of a fraction is the root of the top over the root of the bottom.

2 times 2 is 4.

3 times 3 is 9.

So t − 5/2 is plus or minus this root.

There are two answers: plus and minus.

5 Get t alone

To remove the − 5/2, we add 5/2 on both sides.

We do it in both equations.

Same bottom, so we work with the tops.

3/2 plus 5/2 is 8/2.

That is 4.

Minus 3/2 plus 5/2 is 2/2.

That is 1.

So t = 4 or t = 1.

6 Check

Let's check: put t = 4 into the left side.

We get 0, the same as the right side.

It works!

Now put t = 1 into the left side.

We get 0, the same as the right side.

So both answers are right!

Let's look back.

First: move the 4.

Then: complete the square.

Then: write it as a square.

Then: square root of both sides.

Then: get t alone.

Last: check.