How many solutions x^2 + 2x + 5 = 0 Step by Step

Step by step

The question

Question: find the number of real solutions of x² + 2x + 5 = 0.

x² + 2x + 5 = 0

1 Find a, b and c

We read the numbers: a = 1 is with x², b = 2 is with x, c = 5 is the plain number.

a = 1, b = 2, c = 5

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 = 2² − 4 × 1 × 5

2² is 4.

4 times 1 times 5 is 20.

2² = 2 × 2 = 4

4 × 1 × 5 = 20

Careful!

b is squared first: b² means 2 times 2, not 2.

Careful: not 2 − 20 ✗

So we have 4 take away 20.

= 4 − 20

So the discriminant is −16.

= −16

3 How many solutions

−16 is negative.

No real number times itself is negative.

So the square root of −16 is not a real number.

positive → 2 real solutions

0 → 1 real solution

negative → no real solutions

So x² + 2x + 5 = 0 has no real solutions.

= no real solutions

4 Check

Let's check with the formula.

Under the root we get −16.

x = −2 ± √−162

No real number squared is negative.

So the formula gives no real solutions.

It works!

√(−16): not a real number

Let's look back.

First: find a, b and c.

Then: work out b² − 4ac.

Then: how many solutions.

Last: check.

Answer