Step by step
The question
Question: for which c does x² + 4x + c = 0 have no real solutions?
1 Find a, b and c
We read the numbers.
a = 1 is with x².
b = 4 is with x.
The plain number is the letter c: it is the c we want.
2 Choose the rule
From the discriminant lesson: the sign of b² − 4ac tells how many solutions there are.
We want no real solutions.
So b² − 4ac must be negative, less than 0.
We write it: b² − 4ac < 0.
3 Put the numbers in
We put in the numbers.
The letter c stays as it is: it is the number we want.
Careful!
4² means 4 times 4, not times 2.
4 squared is 16.
4 times 1 times c is 4c.
Now we have 16 − 4c < 0.
We want c by itself.
Two things are stuck to c: times −4 and + 16.
First c was times −4.
Then + 16.
To take them off, we go backwards.
First we remove the + 16.
Then the times −4.
4 Remove the 16
To remove 16, we take away 16.
We do it on both sides, so the sign stays the same.
16 and − 16 make zero.
Cross them out.
0 − 16 = −16.
Now we have −4c < −16.
5 Remove the −4
−4c means −4 times c.
We want just c.
To remove times −4, we divide by −4.
We do it on both sides.
We divide by a negative number.
That turns the sign around.
Less than becomes greater than.
Why?
Look at 2 < 3.
Divide both by −1: we get −2 and −3.
But −2 is more than −3: on the number line, −2 is to the right of −3.
So the sign must turn around.
−4 divided by −4 is 1.
So just c is left.
−16 divided by −4 is 4.
So c is greater than 4.
6 Show it on a number line
7 Check
At c = 4, b² − 4ac is exactly 0.
On the shaded side of 4, it is below 0.
Let's test one number from the answer: c = 5.
It is greater than 4.
b² − 4ac is −4.
−4 is negative.
So there are no real solutions.
It works!
Let's look back.
First: find a, b and c.
Then: choose the rule.
Then: put the numbers in.
Then: remove the 16.
Then: remove the −4.
Then: show it on a number line.
Last: check.