Consider the inequality:

x² − 6x + 8 > 0

It can be solved by following these steps:

1.Equal the polynomial of the first member to zero and the roots of the quadratic equation are obtained.

x² − 6x + 8 = 0

 

2. Represent these values in the real line. Take one point of each interval and evaluate the sign in each:

P(0) = 0² − 6 · 0 + 8 > 0

P(3) = 3² − 6 · 3 + 8 = 17 − 18 < 0

P(5) = 5² − 6 · 5 + 8 = 33 − 30 > 0

3. The solution is defined by the intervals (or the interval) that have the same sign as the polynomial.

S = (-∞, 2) (4, ∞)

 

x² + 2x +1 ≥ 0

x² + 2x +1 = 0

(x + 1)² ≥ 0

As a number squared is always positive the solution is .

Solution
x² + 2x +1 ≥ 0 (x + 1)² ≥ 0
x² + 2x +1 > 0 (x + 1)² > 0
x² + 2x +1 ≤ 0 (x + 1)² ≤ 0 x = − 1
x² + 2x +1 < 0 (x + 1)² < 0

x² + x +1 > 0

x² + x +1 = 0

 

When there are no real roots, give the polynomial any value if:

The sign obtained coincides with the inequality, the solution is .

The obtained sign does not coincide with that of the inequality and thus, has no solution.

Solution
x² + x +1 ≥ 0
x² + x +1 > 0
x² + x +1 ≤ 0
x² + x +1 < 0

 

Example 1

7x² + 21x − 28 < 0

x² +3x − 4 < 0

x² +3x − 4 = 0

P(−6) = (−6)² +3 · (−6)− 4 > 0

P(0) = 0² +3 · 0 − 4 < 0

P(3) = 3² +3 · 3 − 4 > 0

(−4, 1)

The best Maths tutors available
1st lesson free!
Intasar
4.9
4.9 (23 reviews)
Intasar
£42
/h
1st lesson free!
Matthew
5
5 (17 reviews)
Matthew
£25
/h
1st lesson free!
Dr. Kritaphat
4.9
4.9 (6 reviews)
Dr. Kritaphat
£49
/h
1st lesson free!
Paolo
4.9
4.9 (11 reviews)
Paolo
£25
/h
1st lesson free!
Ayush
5
5 (28 reviews)
Ayush
£60
/h
1st lesson free!
Petar
4.9
4.9 (9 reviews)
Petar
£27
/h
1st lesson free!
Rajan
4.9
4.9 (11 reviews)
Rajan
£15
/h
1st lesson free!
Farooq
5
5 (13 reviews)
Farooq
£35
/h
1st lesson free!
Intasar
4.9
4.9 (23 reviews)
Intasar
£42
/h
1st lesson free!
Matthew
5
5 (17 reviews)
Matthew
£25
/h
1st lesson free!
Dr. Kritaphat
4.9
4.9 (6 reviews)
Dr. Kritaphat
£49
/h
1st lesson free!
Paolo
4.9
4.9 (11 reviews)
Paolo
£25
/h
1st lesson free!
Ayush
5
5 (28 reviews)
Ayush
£60
/h
1st lesson free!
Petar
4.9
4.9 (9 reviews)
Petar
£27
/h
1st lesson free!
Rajan
4.9
4.9 (11 reviews)
Rajan
£15
/h
1st lesson free!
Farooq
5
5 (13 reviews)
Farooq
£35
/h
First Lesson Free>

Example 2

−x² + 4x − 7 < 0

x² − 4x + 7 = 0

P(0) = −0² + 4 ·0 − 7 < 0

S =

Example 3

P(−3) = 4 · (−3)² − 16 > 0

P(0) = 4 · 0 ² − 16 < 0

P(3) = 4 · 3 ² − 16 > 0

(-∞, −2] [2, +∞)

Check for the best Maths tutors on Superprof.

Check for outstanding maths tutors near me here.

Get information about maths tuition in the UK.

Need a Maths teacher?

Did you like the article?

1 Star2 Stars3 Stars4 Stars5 Stars 3.00/5 - 2 vote(s)
Loading...

Emma

I am passionate about travelling and currently live and work in Paris. I like to spend my time reading, gardening, running, learning languages and exploring new places.