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Example Questions
Example Question #63 : Linear / Rational / Variable Equations
Three consecutive even numbers add to 42. What is the middle number?
Let = 1st even number,
= 2nd even number, and
= 3rd even number.
Then the equation to solve becomes .
Thus , so the middle number is 14.
Example Question #61 : Equations / Inequalities
Consider the following equation:
Which of the following must be true?
The quantity inside the absolute value brackets must equal either or
, depending on whether the quantity inside the brackets is positive or negative. We therefore have two seperate equations:
To solve the first equation, add 9 to both sides:
Subtract from both sides:
This is the first solution. Now let's look at the second equation. The distributive law gives us:
Add 9 to both sides:
Add to both sides:
Divide both sides by 3:
Therefore, is either 4 or 6.
Statement does NOT have to be true because
can also equal 4.
Statement must be true because both 4 and 6 are positive .
Finally, statement always holds because 4 and 6 are both even.
Example Question #141 : Equations / Inequalities
What property of arithmetic is demonstrated here?
Identity
Reflexive
Commutative
Symmetric
Transitive
Reflexive
The statement expresses the idea that any number is equal to itself. This is the reflexive property of equality.
Example Question #1 : Quadratic Equations
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
–2 or 5
–2 or 4
3 or 4
3 or –5
–3 or 2
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 #141 : Equations / Inequalities
Two consecutive positive multiples of three have a product of 54. What is the sum of the two numbers?
6
9
3
15
12
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 #148 : Equations / Inequalities
Solve 3x2 + 10x = –3
x = –1/9 or –9
x = –2/3 or –2
x = –1/6 or –6
x = –1/3 or –3
x = –4/3 or –1
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 #31 : Quadratic Equations
3x2 – 11x = –10
Which of the following is a valid value for x?
-2
3
None of the other answers
5 / 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 #311 : Algebra
What is the sum of the values of x that satisfy the following equation:
16x – 10(4)x + 16 = 0.
2
5/2
3/2
1
4
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 #5 : How To Find The Solution To A Quadratic Equation
If x > 0, what values of x satisfy the inequality x2 > x?
All positive integers
All positive real numbers
All real numbers greater than one
No values of x satifisfy the inequality
All perfect squares
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.
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