Step by step
The question
Question: for which c does −x² + 6x + c = 0 have exactly one solution?
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 exactly one solution.
So b² − 4ac must be 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!
6² means 6 times 6, not times 2.
6 squared is 36.
4 times −1 times c is −4c.
Taking away a negative is the same as adding.
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 they stay equal.
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, so they stay equal.
4 divided by 4 is 1.
So just c is left.
−36 divided by 4 is −9.
So c = −9!
6 Check
Let's check.
We put c = −9 back in.
b² − 4ac is 36 take away 36: 0.
In the quadratic formula, the square root of 0 is 0.
Plus 0 and minus 0 give the same number: x = 3.
Exactly one solution.
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.
Last: check.