How many solutions -x^2 + 4x - 4 = 0 Step by Step

Step by step

The question

Question: find the number of real solutions of −x² + 4x − 4 = 0.

−x² + 4x − 4 = 0

The x² part is negative.

We change every sign.

0 stays 0.

x² − 4x + 4 = 0

1 Find a, b and c

We read the numbers: a = 1 is with x², b = −4 is with x, c = 4 is the plain number.

a = 1, b = −4, c = 4

2 Work out b² − 4ac

In the formula, b² − 4ac is under the square root.

It is called the discriminant.

Its sign tells us how many solutions there are.

x = −b ± √b² − 4ac2a

We put the numbers in, each negative one in brackets.

b² − 4ac = (−4)² − 4 × 1 × 4

(−4)² is 16.

Minus times minus is plus.

4 times 1 times 4 is 16.

(−4)² = (−4) × (−4) = 16

4 × 1 × 4 = 16

Careful!

b = −4 goes in brackets.

(−4) squared is 16, not minus 16.

Careful: not −4² = −16 ✗

So we have 16 take away 16.

= 16 − 16

So the discriminant is 0.

= 0

3 How many solutions

It is 0.

The square root of 0 is 0.

Plus 0 and minus 0 give the same answer.

So there is just one.

positive → 2 real solutions

0 → 1 real solution

negative → no real solutions

So −x² + 4x − 4 = 0 has 1 real solution.

= 1 real solution

4 Check

Let's check with the formula.

Under the root we get 0, and the square root of 0 is 0.

x = 4 ± √02 = 4 ± 02

Plus 0 and minus 0 give the same number: x = 2.

One solution.

It works!

x = 2 only

Let's look back.

First: find a, b and c.

Then: work out b² − 4ac.

Then: how many solutions.

Last: check.

Answer