All SAT Math Resources
Example Questions
Example Question #1 : Quadratic Equations
In the equation what are the values of ?
None of the other answers
To solve, begin by factoring the equation. We know that in our two factors, one will begin with and one will begin with . Fiddling with different factors of (we are looking for two numbers that, when multiplied by and separately, will add to ), we come to the following:
(If you are unsure, double check by expanding the equation to match the original)
Now, set the each factor equal to 0:
Do the same for the second factor:
Therefore, our two values of are and .
Alternatively, this problem can be solved by plugging each answer choice into the original equation and finding which set of numbers make the equation equate to 0.
Example Question #1 : Quadratic Equations
Solve for :
and
and
and
and
and
and
To solve for , we need to factor the equation to identify its roots, or the values of that make the equation equal .
We can factor out a from each of the terms on the left side of the equation since they have that in common:
Then, we can factor the remaining quadratic portion of the equation. and have a difference of , suggesting that trying and might work:
At this point, we can identify the roots of the equation by setting and and solving each of these equations to yield and , respectively.
Example Question #5 : How To Find The Solution To A Quadratic Equation
If f(x) = -x2 + 6x - 5, then which could be the value of a if f(a) = f(1.5)?
We need to input 1.5 into our function, then we need to input "a" into our function and set these results equal.
f(a) = f(1.5)
f(a) = -(1.5)2 +6(1.5) -5
f(a) = -2.25 + 9 - 5
f(a) = 1.75
-a2 + 6a -5 = 1.75
Multiply both sides by 4, so that we can work with only whole numbers coefficients.
-4a2 + 24a - 20 = 7
Subtract 7 from both sides.
-4a2 + 24a - 27 = 0
Multiply both sides by negative one, just to make more positive coefficients, which are usually easier to work with.
4a2 - 24a + 27 = 0
In order to factor this, we need to mutiply the outer coefficients, which gives us 4(27) = 108. We need to think of two numbers that multiply to give us 108, but add to give us -24. These two numbers are -6 and -18. Now we rewrite the equation as:
4a2 - 6a -18a + 27 = 0
We can now group the first two terms and the last two terms, and then we can factor.
(4a2 - 6a )+(-18a + 27) = 0
2a(2a-3) + -9(2a - 3) = 0
(2a-9)(2a-3) = 0
This means that 2a - 9 =0, or 2a - 3 = 0.
2a - 9 = 0
2a = 9
a = 9/2 = 4.5
2a - 3 = 0
a = 3/2 = 1.5
So a can be either 1.5 or 4.5.
The only answer choice available that could be a is 4.5.
Example Question #162 : Gre Quantitative Reasoning
Solve for x: 2(x + 1)2 – 5 = 27
3 or 4
3 or –5
–2 or 5
–3 or 2
–2 or 4
3 or –5
Quadratic equations generally have two answers. We add 5 to both sides and then divide by 2 to get the quadratic expression on one side of the equation: (x + 1)2 = 16. By taking the square root of both sides we get x + 1 = –4 or x + 1 = 4. Then we subtract 1 from both sides to get x = –5 or x = 3.
Example Question #1 : How To Find The Solution To A Quadratic Equation
Two consecutive positive multiples of three have a product of 54. What is the sum of the two numbers?
12
3
15
9
6
15
Define the variables to be x = first multiple of three and x + 3 = the next consecutive multiple of 3.
Knowing the product of these two numbers is 54 we get the equation x(x + 3) = 54. To solve this quadratic equation we need to multiply it out and set it to zero then factor it. So x2 + 3x – 54 = 0 becomes (x + 9)(x – 6) = 0. Solving for x we get x = –9 or x = 6 and only the positive number is correct. So the two numbers are 6 and 9 and their sum is 15.
Example Question #8 : How To Find The Solution To A Quadratic Equation
Solve 3x2 + 10x = –3
x = –1/3 or –3
x = –1/9 or –9
x = –2/3 or –2
x = –4/3 or –1
x = –1/6 or –6
x = –1/3 or –3
Generally, quadratic equations have two answers.
First, the equations must be put in standard form: 3x2 + 10x + 3 = 0
Second, try to factor the quadratic; however, if that is not possible use the quadratic formula.
Third, check the answer by plugging the answers back into the original equation.
Example Question #1 : Quadratic Equations
3x2 – 11x = –10
Which of the following is a valid value for x?
-2
-5 / 3
None of the other answers
3
5 / 3
5 / 3
Begin by getting our equation into the form Ax2 + BX + C = 0:
3x2 – 11x + 10 = 0
Now, if you factor the left, you can find the answer. Begin by considering the two groups. They will have to begin respectively with 3 and 1 as coefficients for your x value. Likewise, looking at the last element, you can tell that both will have to have a + or –, since the C coefficient is positive. Finally, since the B coefficient is negative, we know that it will have to be –. We know therefore:
(3x – ?)(x – ?)
The potential factors of 10 are: 10, 1; 1, 10; 2, 5; 5, 2
5 and 2 work:
(3x – 5)(x – 2) = 0 because you can FOIL (3x – 5)(x – 2) back into 3x2 – 11x + 10.
Now, the trick remaining is to set each of the factors equal to 0 because if either group is 0, the whole equation will be 0:
3x – 5 = 0 → 3x = 5 → x = 5/3
x – 2 = 0 → x = 2
Therefore, x is either 5 / 3 or 2. The former is presented as an answer.
Example Question #11 : How To Find The Solution To A Quadratic Equation
What is the sum of the values of x that satisfy the following equation:
16x – 10(4)x + 16 = 0.
3/2
2
1
4
5/2
2
The equation we are asked to solve is 16x – 10(4)x + 16 = 0.
Equations of this type can often be "transformed" into other equations, such as linear or quadratic equations, if we rewrite some of the terms.
First, we can notice that 16 = 42. Thus, we can write 16x as (42)x or as (4x)2.
Now, the equation is (4x)2 – 10(4)x + 16 = 0
Let's introduce the variable u, and set it equal 4x. The advantage of this is that it allows us to "transform" the original equation into a quadratic equation.
u2 – 10u + 16 = 0
This is an equation with which we are much more familiar. In order to solve it, we need to factor it and set each factor equal to zero. In order to factor it, we must think of two numbers that multiply to give us 16 and add to give us –10. These two numbers are –8 and –2. Thus, we can factor u2 – 10u + 16 = 0 as follows:
(u – 8)(u – 2) = 0
Next, we set each factor equal to 0.
u – 8 = 0
Add 8.
u = 8
u – 2 = 0
Add 2.
u = 2.
Thus, u must equal 2 or 8. However, we want to find x, not u. Since we defined u as equal to 4x, the equations become:
4x = 2 or 4x = 8
Let's solve 4x = 2 first. We can rewrite 4x as (22)x = 22x, so that the bases are the same.
22x = 2 = 21
2x = 1
x = 1/2
Finally, we will solve 4x = 8. Once again, let's write 4x as 22x. We can also write 8 as 23.
22x = 23
2x = 3
x = 3/2
The original question asks us to find the sum of the values of x that solve the equation. Because x can be 1/2 or 3/2, the sum of 1/2 and 3/2 is 2.
The answer is 2.
Example Question #1 : How To Find The Solution To A Quadratic Equation
If x > 0, what values of x satisfy the inequality x2 > x?
No values of x satifisfy the inequality
All positive integers
All perfect squares
All real numbers greater than one
All positive real numbers
All real numbers greater than one
There are two values where x2 = x, namely x = 0 and x = 1. All values between 0 and 1 get smaller after squaring. All values greater than 1 get larger after squaring.
Example Question #11 : Quadratic Equations
Let f(x) = 2x2 – 4x + 1 and g(x) = (x2 + 16)(1/2). If k is a negative number such that f(k) = 31, then what is the value of (f(g(k))?
-81
-35
5
31
25
31
In order to find the value of f(g(k)), we will first need to find k. We are told that f(k) = 31, so we can write an expression for f(k) and solve for k.
f(x) = 2x2 – 4x + 1
f(k) = 2k2 – 4k + 1 = 31
Subtract 31 from both sides.
2k2 – 4k – 30 = 0
Divide both sides by 2.
k2 – 2k – 15 = 0
Now, we can factor this by thinking of two numbers that multiply to give –15 and add to give –2. These two numbers are –5 and 3.
k2 –2k – 15 = (k – 5)(k + 3) = 0
We can set each factor equal to 0 to find the values for k.
k – 5 = 0
Add 5 to both sides.
k = 5
Now we set k + 3 = 0.
Subtract 3 from both sides.
k = –3
This means that k could be either 5 or –3. However, we are told that k is a negative number, which means k = –3.
Finally, we can evaluate the expression f(g(–3)). First we need to find g(–3).
g(x) = (x2 + 16)(1/2)
g(–3) = ((–3)2 + 16)(1/2)
= (9 + 16)(1/2)
= 25(1/2)
Raising something to the one-half power is the same as taking the square root.
25(1/2) = 5
Now that we know g(–3) = 5, we must find f(5).
f(5) = 2(5)2 – 4(5) + 1
= 2(25) – 20 + 1 = 31
The answer is 31.