All SAT Math Resources
Example Questions
Example Question #191 : Algebra
Give the solution set of the inequality
In an absolute value inequality, the absolute value expression must be isolated on one side first. We can do this by subtracting 21 from both sides:
This can be rewritten as the three-part inequality
Subtract 14 from all three expressions:
Divide all three expressions by , reversing the inequality symbols since you are dividing by a negative number:
In interval notation, this is .
Example Question #21 : How To Find A Solution Set
Give the solution set of the inequality
The inequality has no solution.
In an absolute value inequality, the absolute value expression must be isolated on one side first. We can di this by first subtracting 42 from both sides:
Divide by , reversing the direction of the inequality symbol since we are dividing by a negative number:
This inequality can be rewritten as the compound inequality
or
Solve each simple inequality separately.
Subtract 19 from both sides:
Divide by , remembering to reverse the symbol:
In interval notation, this is .
Carry out the same steps on the other simple inequality:
In interval notation, this is .
Since the two simple inequalities are connected by an "or", their individual solution sets are connected by a union; the solution set is
.
Example Question #22 : How To Find A Solution Set
Below is a table of earnings from playing blackjack.
Find the equation of depreciation.
In general, an equation of depreciation looks like the following.
, where is the starting amount, is the common ratio, and is time.
For us , , and .
Example Question #191 : Algebra
The above displays a scatterplot, where the red line is the line of best fit. The line of best fit equation is . What point on the scatterplot is the furthest from the line of best fit?
To answer this question, we should create a table of values and compare. For calculating the difference between the y value and the y value of the line of best, take the absolute value since it won't make since if it is negative.
From our table of values, we can see that the point furthest from the line of best fit is .
Example Question #24 : How To Find A Solution Set
Find the solutions to
First step is to set it equal to zero
Now factor out an
Set up the equations, and solve for .
Example Question #192 : Algebra
Factor the following equation.
x2 – 16
(x + 4)(x + 4)
(x)(x – 4)
(x + 4)(x – 4)
(x – 4)(x – 4)
(x2)(4 – 2)
(x + 4)(x – 4)
The correct answer is (x + 4)(x – 4)
We neen to factor x2 – 16 to solve. We know that each parenthesis will contain an x to make the x2. We know that the root of 16 is 4 and since it is negative and no value of x is present we can tell that one 4 must be positive and the other negative. If we work it from the multiple choice answers we will see that when multiplying it out we get x2 + 4x – 4x – 16. 4x – 4x cancels out and we are left with our answer.
Example Question #3 : How To Factor An Equation
If x3 – y3 = 30, and x2 + xy + y2 = 6, then what is x2 – 2xy + y2?
180
25
cannot be determined
5
24
25
First, let's factor x3 – y3 using the formula for difference of cubes.
x3 – y3 = (x – y)(x2 + xy + y2)
We are told that x2 + xy + y2 = 6. Thus, we can substitute 6 into the above equation and solve for x – y.
(x - y)(6) = 30.
Divide both sides by 6.
x – y = 5.
The original questions asks us to find x2 – 2xy + y2. Notice that if we factor x2 – 2xy + y2 using the formula for perfect squares, we obtain the following:
x2 – 2xy + y2 = (x – y)2.
Since we know that (x – y) = 5, (x – y)2 must equal 52, or 25.
Thus, x2 – 2xy + y2 = 25.
The answer is 25.
Example Question #4 : How To Factor An Equation
if x – y = 4 and x2 – y = 34, what is x?
12
6
9
10
15
6
This can be solved by substitution and factoring.
x2 – y = 34 can be written as y = x2 – 34 and substituted into the other equation: x – y = 4 which leads to x – x2 + 34 = 4 which can be written as x2 – x – 30 = 0.
x2 – x – 30 = 0 can be factored to (x – 6)(x + 5) = 0 so x = 6 and –5 and because only 6 is a possible answer, it is the correct choice.
Example Question #5 : How To Factor An Equation
If x2 + 2ax + 81 = 0. When a = 9, what is the value of x?
3
0
–9
–18
9
–9
When a = 9, then x2 + 2ax + 81 = 0 becomes
x2 + 18x + 81 = 0.
This equation can be factored as (x + 9)2 = 0.
Therefore when a = 9, x = –9.
Example Question #1 : Factoring Equations
If f(x) has roots at x = –1, 0 and 2, which of the following could be the equation for f(x)?
f(x) = x2 – x – 2
f(x) = x4 + x3 – 2x2
f(x) = x2 + x – 2
f(x) = x3 – x2 – 2x
f(x) = x3 – x2 + 2x
f(x) = x3 – x2 – 2x
In general, if a function has a root at x = r, then (x – r) must be a factor of f(x). In this problem, we are told that f(x) has roots at –1, 0 and 2. This means that the following are all factors of f(x):
(x – (–1)) = x + 1
(x – 0) = x
and (x – 2).
This means that we must look for an equation for f(x) that has the factors (x + 1), x, and (x – 2).
We can immediately eliminate the function f(x) = x2 + x – 2, because we cannot factor an x out of this polynomial. For the same reason, we can eliminate f(x) = x2 – x – 2.
Let's look at the function f(x) = x3 – x2 + 2x. When we factor this, we are left with x(x2 – x + 2). We cannot factor this polynomial any further. Thus, x + 1 and x – 2 are not factors of this function, so it can't be the answer.
Next, let's examine f(x) = x4 + x3 – 2x2 .
We can factor out x2.
x2 (x2 + x – 2)
When we factor x2 + x – 2, we will get (x + 2)(x – 1). These factors are not the same as x – 2 and x + 1.
The only function with the right factors is f(x) = x3 – x2 – 2x.
When we factor out an x, we get (x2 – x – 2), which then factors into (x – 2)(x + 1). Thus, this function has all of the factors we need.
The answer is f(x) = x3 – x2 – 2x.