All Algebra II Resources
Example Questions
Example Question #181 : Quadratic Equations And Inequalities
Find the roots of the following equation:
A polynominal with the form multiplies out to , so this leaves two equations:
Factors of 12 are 2, 6, and 3, 4. In this case, because the product is negative, one root is positive and one is negative. Because the sum is negative, the positive root must have a lower absolute value than the negative root. This leaves us with two possibilities: 2, -6 and 3, -4. Plugging into the sum equation shows the roots to be 3, -4. A check into the original polynominal shows
Example Question #21 : Solving Quadratic Equations
Solve for .
When asked to solve for , we are really searching for the roots/-intercepts of the equation.
In this particular case, our function is already factored for us leaving us with only a few steps to complete the problem.
Our first step is to set each term equal to , leaving us with...
and
The next step is to use our knowledge of order of operations to simply solve for for each of the above equations...
Subtract from both sides
Divide by
Answer #1, our first root/-intercept
Add to both sides
Answer # 2, our second root/-intercept
Example Question #1484 : Algebra Ii
Solve for .
This problem requires simplification, order of operations, and knowledge of square roots.
Our goal is to isolate/solve for
Divide by on both sides
Square root both sides
**Remember: DO NOT FORGET THAT WHEN WE TAKE A SQUARE ROOT, WE GET A PLUS/MINUS ANSWER.
or
and
Example Question #3 : Binomials
.
Example Question #2 : Solving Equations And Inequallities
Give all real solutions of the following equation:
The equation has no real solutions.
By substituting - and, subsequently, this can be rewritten as a quadratic equation, and solved as such:
We are looking to factor the quadratic expression as , replacing the two question marks with integers with product 36 and sum ; these integers are .
Substitute back:
These factors can themselves be factored as the difference of squares:
Set each factor to zero and solve:
The solution set is .
Example Question #22 : How To Find The Solution To A Quadratic Equation
Solve for x.
x = 4
Cannot be factored by grouping.
x = –6, –4
x = 6, 4
x = –8, –2
x = –8, –2
1) Quadratics must be set equal to zero in order to be solved. To do so in this equation, the "8" has to wind up on the left side and combine with any other lone integers. So, multiply out the terms in order to make it possible for the "8" to be added to the other number.
Then combine like terms.
2) Now factor.
1 + 16 = 17
4 + 4 = 8
2 + 8 = 10
3) Pull out common factors, "x" and "8," respectively.
4) Pull out "(x+2)" from both terms.
x = –8, –2
Example Question #4 : How To Factor The Quadratic Equation
Solve for x.
This is a factoring problem, so we need to get all of the variables on one side and set the equation equal to zero. To do this we subtract 128 from both sides to get .
We then notice that all four numbers are divisible by four, so we can simplify the expression to .
Think of the equation in this format to help with the following explanation.
We must then factor to find the solutions for x. To do this we must make a factor tree of c (which is 32 in this case) to find the possible solutions. The possible numbers are 1 * 32, 2 * 16, and 4 * 8.
Since c is negative, we know that our factoring will produce a positive and negative number.
We then look at b to see if the greater number will be positive or negative. Since b is positive, we know that the greater number from our factoring tree will be positive.
We then use addition and subtraction with the factoring tree to find the numbers that add together to equal b. Remember that the greater number is positive and the lesser number is negative in this example.
Positive 8 and negative 4 equal b. We then plug our numbers into the factored form of .
We know that anything multiplied by 0 is equal to 0, so we plug in the numbers for x which make each equation equal to 0. In this case .
Example Question #181 : Quadratic Equations And Inequalities
Solve for x.
x = –2/3, –3
x = –12, –1
x = –4, –1
x = 4, 1
x = –9, –2
x = –2/3, –3
1) Combine like terms and simplify.
No further simplification is possible. The first term has a coefficient that can't be factored away. FOIL requires that all terms be multiplied by each other at some point, so the presence of the coefficient has to be reflected in every step of the factoring.
2) Practically speaking, that means we add an extra step. Multiply the coefficient of the first term by the last term before factoring.
3 * 6 = 18
Factors of 18 include:
1 + 18 = 19
2 + 9 = 11
3) Now pull out the common factor in each of the pairs, "3x" from the first two and "2" from the second two.
4) Pull out the "(x+3)" from both terms.
5) Set both parts equal to zero and solve.
3x + 2 = 0, x = –2/3
x + 3 = 0, x = –3
Example Question #32 : Finding Roots
Find the roots of .
First, you need to factor the expression. This takes a little bit of trial and error, but even though the leading coefficient is not 1, there's only one way to get 2 (2 x 1) so that is helpful to remember!
Factored, the expression is .
Once you have those expressions, you can set them equal to 0 to get the roots, or solutions to the equation.
When you set .
And when you set
.
Those are your two roots!
Example Question #32 : Solving Quadratic Equations
Solve
You need to first factor the equation. Then you willbe able to set those expressions equal to 0 and find your roots. There is only one way to get your leading coefficient of 3 (3 x 1) so that's helpful when figuring it out.
In the form remember you want a factor of that when mulitplied by the factor of and added together will give you the value of .
Factored, it comes out to:
.
When you set those expressions equal to 0, you get
.
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