MAT1510
PRECALCULUS MATHS 1
Jan-Feb 2016
QUESTION 1
Solve each of the following for 𝑥
1.1 |−2𝑥 − 3| ≤ 1
There are several ways of solving this problem; we will highlight here only two methods.
Method 1
The equation involved the | | modulus sign. To remove the modulus sign, we square both
sides of the equation to get,
|−2𝑥 − 3|2 ≤ 12
(−2𝑥 − 3)(−2𝑥 − 3) ≤ 1
4𝑥 2 + 6𝑥 + 6𝑥 + 9 ≤ 1
4𝑥 2 + 12𝑥 + 8 ≤ 0
𝑥 2 + 3𝑥 + 2 ≤ 0
(𝑥 + 2)(𝑥 + 1) ≤ 0
-2 -1
x
So the solution set for the above equation is
−𝟐 ≤ 𝒙 ≤ −𝟏
Method 2
Since the term −2𝑥 − 3is inside the modulus sign, it can be nnegative or positive.
Taking it to be positive we have
−2𝑥 − 3 ≤ 1
−2𝑥 ≤ 4
Dividing both sides by -2 (and note the sign change due to division by negative), we have
𝑥 ≥ −2
Taking it to be negative we have
−(−2𝑥 − 3) ≤ 1
2𝑥 + 3 ≤ 1
2𝑥 ≤ −2
𝑥 ≤ −1
Combining the two solutions, therefore the full solution set is. −𝟐 ≤ 𝒙 ≤ −𝟏
Jan-Feb 2016 Suggested Solutions
by T. Chinaka
PRECALCULUS MATHS 1
Jan-Feb 2016
QUESTION 1
Solve each of the following for 𝑥
1.1 |−2𝑥 − 3| ≤ 1
There are several ways of solving this problem; we will highlight here only two methods.
Method 1
The equation involved the | | modulus sign. To remove the modulus sign, we square both
sides of the equation to get,
|−2𝑥 − 3|2 ≤ 12
(−2𝑥 − 3)(−2𝑥 − 3) ≤ 1
4𝑥 2 + 6𝑥 + 6𝑥 + 9 ≤ 1
4𝑥 2 + 12𝑥 + 8 ≤ 0
𝑥 2 + 3𝑥 + 2 ≤ 0
(𝑥 + 2)(𝑥 + 1) ≤ 0
-2 -1
x
So the solution set for the above equation is
−𝟐 ≤ 𝒙 ≤ −𝟏
Method 2
Since the term −2𝑥 − 3is inside the modulus sign, it can be nnegative or positive.
Taking it to be positive we have
−2𝑥 − 3 ≤ 1
−2𝑥 ≤ 4
Dividing both sides by -2 (and note the sign change due to division by negative), we have
𝑥 ≥ −2
Taking it to be negative we have
−(−2𝑥 − 3) ≤ 1
2𝑥 + 3 ≤ 1
2𝑥 ≤ −2
𝑥 ≤ −1
Combining the two solutions, therefore the full solution set is. −𝟐 ≤ 𝒙 ≤ −𝟏
Jan-Feb 2016 Suggested Solutions
by T. Chinaka
