Worked examples
The question
Question: find the number of real solutions of x² + 4x + 3 = 0.
x2 + 4x + 3 = 0
1 Find a, b and c
We read the numbers: a = 1 is with x², b = 4 is with x, c = 3 is the plain number.
a = 1, b = 4, c = 3
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 = 42 − 4 × 1 × 3
4² is 16.
4 times 1 times 3 is 12.
4² = 4 × 4 = 16
4 × 1 × 3 = 12
Careful!
b is squared first: b² means 4 times 4, not 4.
Careful: not 4 − 12 ✗
So we have 16 take away 12.
= 16 − 12
So the discriminant is 4.
= 4
3 How many solutions
4 is positive.
The square root of a positive number is a real number.
The plus or minus gives two different answers.
positive → 2 real solutions
0 → 1 real solution
negative → no real solutions
So x² + 4x + 3 = 0 has 2 real solutions.
= 2 real solutions
4 Check
Let's check with the formula.
Under the root we get 4.
Its square root is 2.
x = −4 ± 22
The plus gives −1, the minus gives −3.
Two different solutions.
It works!
x = −1 or −3
Let's look back.
First: find a, b and c.
Then: work out b² − 4ac.
Then: how many solutions.
Last: check.
Answer = 2 real solutions
▶ Watch it solved
The question
Question: find the number of real solutions of 2x² + 9x − 5 = 0.
2x2 + 9x − 5 = 0
1 Find a, b and c
We read the numbers: a = 2 is with x², b = 9 is with x, c = −5 is the plain number.
a = 2, b = 9, 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 = 92 − 4 × 2 × (−5)
9² is 81.
4 times 2 times −5 is −40.
9² = 9 × 9 = 81
4 × 2 × (−5) = −40
Careful!
c is negative, so 4ac is negative.
Taking away −40 is adding 40.
Careful: not 81 − 40 ✗
So we have 81 take away −40.
= 81 − (−40)
Taking away a negative is the same as adding.
81 − (−40) = 81 + 40
So the discriminant is 121.
= 121
3 How many solutions
121 is positive.
The square root of a positive number is a real number.
The plus or minus gives two different answers.
positive → 2 real solutions
0 → 1 real solution
negative → no real solutions
So 2x² + 9x − 5 = 0 has 2 real solutions.
= 2 real solutions
4 Check
Let's check with the formula.
Under the root we get 121.
Its square root is 11.
x = −9 ± 114
The plus gives 1/2, the minus gives −5.
Two different solutions.
It works!
x = 12 or −5
Let's look back.
First: find a, b and c.
Then: work out b² − 4ac.
Then: how many solutions.
Last: check.
Answer = 2 real solutions
▶ Watch it solved
The question
Question: find the number of real solutions of 2x² = 3x + 5.
2x2 = 3x + 5
1 Get 0 on one side
We want 0 on the right side.
So we take away 3x + 5 on both sides.
We do it on both sides, so they stay equal.
2x2 − 3x − 5 = 3x + 5 − 3x − 5
Now we have 2x² − 3x − 5 = 0.
2x2 − 3x − 5 = 0
2 Find a, b and c
We read the numbers: a = 2 is with x², b = −3 is with x, c = −5 is the plain number.
a = 2, b = −3, c = −5
3 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 = (−3)2 − 4 × 2 × (−5)
(−3)² is 9.
Minus times minus is plus.
4 times 2 times −5 is −40.
(−3)² = (−3) × (−3) = 9
4 × 2 × (−5) = −40
Careful!
b = −3 goes in brackets.
(−3) squared is 9, not minus 9.
Careful: not −3² = −9 ✗
So we have 9 take away −40.
= 9 − (−40)
Taking away a negative is the same as adding.
9 − (−40) = 9 + 40
So the discriminant is 49.
= 49
4 How many solutions
49 is positive.
The square root of a positive number is a real number.
The plus or minus gives two different answers.
positive → 2 real solutions
0 → 1 real solution
negative → no real solutions
So 2x² = 3x + 5 has 2 real solutions.
= 2 real solutions
5 Check
Let's check with the formula.
Under the root we get 49.
Its square root is 7.
x = 3 ± 74
The plus gives 5/2, the minus gives −1.
Two different solutions.
It works!
x = 52 or −1
Let's look back.
First: get 0 on one side.
Then: find a, b and c.
Then: work out b² − 4ac.
Then: how many solutions.
Last: check.
Answer = 2 real solutions
▶ Watch it solved
