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Example Questions
Example Question #3 : Solving Quadratic Equations
Solve the following quadratic equation:
When solving a quadratic equation, the first thing to look for is whether or not it can be factored, as this is most often the easiest and fastest method if the quadratic can in fact be factored. We can see that each of the terms in the given equation have a common factor of 3, so it will be easier to factor the quadratic if we first factor out the 3:
Now we're left with a polynomial where we need to find two numbers whose product is -28 and whose sum is -3. Thinking about the factors of 28, we can see that 4 and 7 will yield -3 if 7 is negative and 4 is positive, so we now have our factorization:
Example Question #4 : Solving Quadratic Equations
Find the root(s) of the following quadratic equation?
To find the roots of an equation in the form , you use the quadratic formula
.
In our case, we have .
This gives us which simplifies to
Example Question #5 : Solving Quadratic Equations
Given the function , find a possible root for this quadratic.
Factorize and set this equation equal to zero.
The answer is one of the possible choices.
Example Question #11 : Solving Quadratic Equations
Solve the quadratic equation for .
None of the other answers.
None of the other answers.
There are two solutions; .
We proceed as follows.
Add to both sides.
Take the square root of both sides, remember to introduce plus/minus on the right side since you are introducing a square root into your work.
Add to both sides.
Example Question #12 : Solving Quadratic Equations
Solve the quadratic equation for .
Use the quadratic formula.
None of the other answers.
For any quadtratic equation of the form , the quadratic formula is
Plugging in our given values we have:
Example Question #6 : Solve A Quadratic Equation
Find the roots of the equation.
Use either the quadratic formula or factoring to solve the quadratic equation.
Using factoring, we want to find which factors of six when multiplied with the factors of two and then added together result in negative one.
Using the quadratic formula,
let
Example Question #7 : Solve A Quadratic Equation
Solve .
To solve this equation, use trial and error to factor it. Since the leading coefficient is , there is only one way to get , so that is helpful reminder. Once it's properly factored, you get: . Then, set both of those expressions equal to to get your roots: .
Example Question #1 : Solving Polynomial Equations
Solve the following polynomial for by factoring:
The polynomial in the problem is given as follows:
Factoring this polynomial, we would get an expression of the form:
So we need to determine what a and b are. We know we need two factors that when multiplied equal -12, and when added equal -1. If we consider 2 and 6, we could get -12 but could not arrange them in any way that would make their sum equal to -1. We then look at 3 and 4, whose product can be -12 is one of them is negative, and whose sum can be -1 if -4 is added to 3. This tells us that the 4 must be the negative factor and the 3 must be the positive factor, so we get the following:
Example Question #2 : Solving Polynomial Equations
Solve the following polynomial for by factoring:
Factoring our polynomial, we can see we will have 2x and x at the beginning of each factor, while we need to find two numbers whose product is 15 and whose sum when multiplied by our leading terms and added is -11x. This gives us the following factorization:
So now we can set each term equal to 0 and solve for our two values of x:
Example Question #8 : Solving Polynomial Equations
Find the root(s) for the function:
The function is in the form , and can be factorized.
Determine the values of and .
Substitute this into the formula.
Set to find all roots.
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