Step by step
The question
Question: for which c does x² − 6x + c = 0 have two real solutions?
1 Find a, b and c
We read the numbers.
a = 1 is with x².
b = −6 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 two real solutions.
So b² − 4ac must be positive, more than 0.
We write it: b² − 4ac > 0.
3 Put the numbers in
We put in the numbers, each negative one in brackets.
The letter c stays as it is: it is the number we want.
Careful!
(−6)² means (−6) times (−6), not times 2.
(−6) squared is 36.
Minus times minus is plus.
4 times 1 times c is 4c.
Now we have 36 − 4c > 0.
We want c by itself.
Two things are stuck to c: times −4 and + 36.
First c was times −4.
Then + 36.
To take them off, we go backwards.
First we remove the + 36.
Then the times −4.
4 Remove the 36
To remove 36, we take away 36.
We do it on both sides, so the sign stays the same.
36 and − 36 make zero.
Cross them out.
0 − 36 = −36.
Now we have −4c > −36.
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.
Greater than becomes less 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.
−36 divided by −4 is 9.
So c is less than 9.
6 Show it on a number line
7 Check
At c = 9, b² − 4ac is exactly 0.
On the shaded side of 9, it is above 0.
Let's test one number from the answer: c = 8.
It is less than 9.
b² − 4ac is 4.
4 is positive.
So there are two 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 36.
Then: remove the −4.
Then: show it on a number line.
Last: check.