a Rewrite the inequality so that all variable terms are on one side of the inequality and all constants are on the other side.
B
b Calculate ( b2 )^2.
C
c Rewrite the inequality so that all variable terms are on one side of the inequality and all constants are on the other side.
D
d Rewrite the equation so that all variable terms are on one side of the equation and all constants are on the other side.
A
a - 1 ≤ x ≤ 3
B
b x=- 2± sqrt(7)
C
c All real numbers.
D
d x=11/2 or x=- 5/2
Practice makes perfect
a We want to solve the quadratic inequality by completing the square. Let's start by rewriting the inequality so all terms with x are on one side of the inequality and all constants are on the other side.
x^2-2x-3 ≤ 0
⇕
x^2-2x ≤ 3
In a quadratic expression, b is the linear coefficient. For the inequality above, we have that b=- 2. Let's now calculate ( b2 )^2.
Now we can take the square root of the inequality. Remember that we do not know the sign of x. Thus, to consider both possibilities we need to use the absolute value.
Now, we can create a compound inequality by removing the absolute value. In this case, the solution set is any number less than or equal to 2 away from the midpoint in the positive direction and any number less than or equal to 2 away from the midpoint in the negative direction.
Absolute Value Inequality:& |x-1| ≤ 2
Compound Inequality:& - 2≤ x-1 ≤ 2
We can split this compound inequality into two cases, one where x-1 is greater than or equal to -2 and one where x-1 is less than or equal to 2.
x-1 ≥ - 2 and x-1 ≤ 2
Let's isolate x in both of these cases before graphing the solution set.
This inequality tells us that all values greater than or equal to - 1 will satisfy the inequality.
Solution Set
The solution to this type of compound inequality is the overlap of the solution sets. Let's recombine our cases back into one compound inequality.
First Solution Set:& x ≤ 3
Second Solution Set:& - 1 ≤ x
Intersecting Solution Set:& - 1 ≤ x ≤ 3
b We want to solve the quadratic equation by completing the square. Note that all terms with x are on one side of the equation.
x^2+4x=3In a quadratic expression b is the linear coefficient. For the equation above, we have that b=4. Let's now calculate ( b2 )^2.
Both x=- 2+ sqrt(7) and x=- 2- sqrt(7) are solutions of the equation.
c We want to solve the quadratic inequality by completing the square. Let's start by rewriting the inequality so all terms with x are on one side of the inequality and all constants are on the other side.
x^2+12x+39 > 0
⇕
x^2+12x > - 39In a quadratic expression, b is the linear coefficient. For the inequality above we have that b=12. Let's now calculate ( b2 )^2.
Now we would like to take the square root of this inequality. But we cannot! Notice that the right-hand side of the inequality is negative. We can take the square root of an inequality only when both sides are positive. How else can we solve this inequality?
(x+6)^2 > - 3
Recall that a square is always non-negative. Thus, regardless of the number we substitute for x, the square of it, and the square of x+6, will always be non-negative. Therefore, it will be greater than - 3. Since any value of x satisfies the inequality, the solution is all real numbers.
d We want to solve the quadratic equation by completing the square. To do so we will start by rewriting the equation so all terms with x are on one side of the equation and all constants are on the other side.
x^2-3x-13.75=0
⇕
x^2-3x=13.75In a quadratic expression, b is the linear coefficient. For the equation above we have that b=- 3. Let's now calculate ( b2 )^2.
