Worked examples
The question
Question: for which c does x² + 6x + c = 0 have exactly one solution?
x2 + 6x + c = 0
we want: 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.
a = 1, b = 6, c = ?
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.
positive → 2 real solutions
0 → 1 real solution
negative → no real solutions
We write it: b² − 4ac = 0.
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.
62 − 4 × 1 × c = 0
Careful!
6² means 6 times 6, not times 2.
Careful: not 6² = 12 ✗
6 squared is 36.
4 times 1 times c is 4c.
6² = 6 × 6 = 36
4 × 1 × c = 4c
Now we have 36 − 4c = 0.
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.
c → × −4 → + 36
take off: − 36, then ÷ −4
4 Remove the 36
To remove 36, we take away 36.
We do it on both sides, so they stay equal.
36 − 4c − 36 = 0 − 36
36 and − 36 make zero.
Cross them out.
0 − 36 = −36.
0 − 36 = −36
Now we have −4c = −36.
−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.
−4c−4 = −36−4
−4 divided by −4 is 1.
So just c is left.
−36 divided by −4 is 9.
−36−4 = 9
So c = 9!
c = 9
6 Check
Let's check.
We put c = 9 back in.
x2 + 6x + 9 = 0
b² − 4ac is 36 take away 36: 0.
b² − 4ac = 6² − 4 × 1 × 9 = 36 − 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!
x = −6 ± 02 = −3 only
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.
▶ Watch it solved
The question
Question: for which c does 2x² − 8x + c = 0 have exactly one solution?
2x2 − 8x + c = 0
we want: exactly one solution
1 Find a, b and c
We read the numbers.
a = 2 is with x².
b = −8 is with x.
The plain number is the letter c: it is the c we want.
a = 2, b = −8, c = ?
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.
positive → 2 real solutions
0 → 1 real solution
negative → no real solutions
We write it: b² − 4ac = 0.
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.
(−8)2 − 4 × 2 × c = 0
Careful!
(−8)² means (−8) times (−8), not times 2.
Careful: not (−8)² = −16 ✗
(−8) squared is 64.
Minus times minus is plus.
4 times 2 times c is 8c.
(−8)² = (−8) × (−8) = 64
4 × 2 × c = 8c
Now we have 64 − 8c = 0.
64 − 8c = 0
We want c by itself.
Two things are stuck to c: times −8 and + 64.
First c was times −8.
Then + 64.
To take them off, we go backwards.
First we remove the + 64.
Then the times −8.
c → × −8 → + 64
take off: − 64, then ÷ −8
4 Remove the 64
To remove 64, we take away 64.
We do it on both sides, so they stay equal.
64 − 8c − 64 = 0 − 64
64 and − 64 make zero.
Cross them out.
0 − 64 = −64.
0 − 64 = −64
Now we have −8c = −64.
−8c = −64
5 Remove the −8
−8c means −8 times c.
We want just c.
To remove times −8, we divide by −8.
We do it on both sides, so they stay equal.
−8c−8 = −64−8
−8 divided by −8 is 1.
So just c is left.
−64 divided by −8 is 8.
−64−8 = 8
So c = 8!
c = 8
6 Check
Let's check.
We put c = 8 back in.
2x2 − 8x + 8 = 0
b² − 4ac is 64 take away 64: 0.
b² − 4ac = (−8)² − 4 × 2 × 8 = 64 − 64 = 0
In the quadratic formula, the square root of 0 is 0.
Plus 0 and minus 0 give the same number: x = 2.
Exactly one solution.
It works!
x = 8 ± 04 = 2 only
Let's look back.
First: find a, b and c.
Then: choose the rule.
Then: put the numbers in.
Then: remove the 64.
Then: remove the −8.
Last: check.
▶ Watch it solved
The question
Question: for which c does −x² + 6x + c = 0 have exactly one solution?
−x2 + 6x + c = 0
we want: 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.
a = −1, b = 6, c = ?
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.
positive → 2 real solutions
0 → 1 real solution
negative → no real solutions
We write it: b² − 4ac = 0.
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.
62 − 4 × (−1) × c = 0
Careful!
6² means 6 times 6, not times 2.
Careful: not 6² = 12 ✗
6 squared is 36.
4 times −1 times c is −4c.
6² = 6 × 6 = 36
4 × (−1) × c = −4c
Taking away a negative is the same as adding.
36 − (−4c) = 36 + 4c
Now we have 36 + 4c = 0.
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.
c → × 4 → + 36
take off: − 36, then ÷ 4
4 Remove the 36
To remove 36, we take away 36.
We do it on both sides, so they stay equal.
36 + 4c − 36 = 0 − 36
36 and − 36 make zero.
Cross them out.
0 − 36 = −36.
0 − 36 = −36
Now we have 4c = −36.
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.
4c4 = −364
4 divided by 4 is 1.
So just c is left.
−36 divided by 4 is −9.
−364 = −9
So c = −9!
c = −9
6 Check
Let's check.
We put c = −9 back in.
−x2 + 6x − 9 = 0
b² − 4ac is 36 take away 36: 0.
b² − 4ac = 6² − 4 × (−1) × (−9) = 36 − 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!
x = −6 ± 0-2 = 3 only
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.
▶ Watch it solved
