All Algebra II Resources
Example Questions
Example Question #11 : Finding Roots
FInd the roots for
Notice in this question there are only two terms, the exponent value and the constant value. There is also the negative sign between the two. When we look at each number we see that each are a perfect square. Due to the negative sign between the two, this type of quadratic expression can also be written as a difference of squares. We look at the exponential term and see it is
The perfect square factors of this term are and .
Now we look at the constant term
The perfect square factors of this term are and
Now to combine these into the binomial factor form we need to remember it is the difference of perfect squares meaning we will have one subtraction sign and one adding sign, so we get the following:
From here we solve each binomial for x. To do this we set each binomal to zero and solve for x.
Example Question #12 : Finding Roots
Find the roots for
Notice in this question there are only two terms, the exponent value and the constant value. There is also the negative sign between the two. When we look at each number we see that each are a perfect square. Due to the negative sign between the two, this type of quadratic expression can also be written as a difference of squares. We look at the exponential term and see it is
The perfect square factors of this term are and .
Now we look at the constant term
The perfect square factors of this term are and
Now to combine these into the binomial factor form we need to remember it is the difference of perfect squares meaning we will have one subtraction sign and one adding sign, so we get the following:
From here we solve each binomial for x. To do this we set each binomal to zero and solve for x.
Example Question #13 : Finding Roots
Find the roots for
Notice in this question there are only two terms, the exponent value and the constant value. There is also the negative sign between the two. When we look at each number we see that each are a perfect square. Due to the negative sign between the two, this type of quadratic expression can also be written as a difference of squares. We look at the exponential term and see it is
The perfect square factors of this term are and .
Now we look at the constant term
The perfect square factors of this term are and
Now to combine these into the binomial factor form we need to remember it is the difference of perfect squares meaning we will have one subtraction sign and one adding sign, so we get the following:
From here we solve each binomial for x. To do this we set each binomal to zero and solve for x.
Example Question #14 : Finding Roots
Find the roots for
Notice in this question there are only two terms, the exponent value and the constant value. There is also the negative sign between the two. When we look at each number we see that each are a perfect square. Due to the negative sign between the two, this type of quadratic expression can also be written as a difference of squares. We look at the exponential term and see it is
The perfect square factors of this term are and .
Now we look at the constant term
The perfect square factors of this term are and .
Now to combine these into the binomial factor form we need to remember it is the difference of perfect squares meaning we will have one subtraction sign and one adding sign, so we get the following:
From here we solve each binomial for x. To do this we set each binomal to zero and solve for x.
Example Question #15 : Finding Roots
Find the roots of .
Notice in this question there are only two terms, the exponent value and the constant value. There is also the negative sign between the two. When we look at each term we see that each is a perfect square. Due to the negative sign between the two, this type of quadratic expression can also be written as a difference of squares. We look at the exponential term and see it is
The perfect square factors of this term are and .
Now we look at the constant term
The perfect square factors of this term are and
Now to combine these into the binomial factor form we need to remember it is the difference of perfect squares meaning we will have one subtraction sign and one adding sign, so we get the following:
From here we solve each binomial for x. To do this we set each binomal to zero and solve for x.
Example Question #16 : Finding Roots
Find the roots of .
Notice in this question there are only two terms, the exponent value and the constant value. There is also the negative sign between the two. When we look at each term we see that each is a perfect square. Due to the negative sign between the two, this type of quadratic expression can also be written as a difference of squares. We look at the exponential term and see it is
The perfect square factors of this term are and .
Now we look at the constant term
The perfecct square factors of this term are and .
Now to combine these into the binomial factor form we need to remember it is the difference of perfect squares meaning we will have one subtraction sign and one adding sign, so we get the following:
From here we solve each binomial for x. To do this we set each binomal to zero and solve for x.
Example Question #12 : Finding Roots
Solve for :
To solve the quadratic equation, , first set the equation equal to zero and then factor the quadratic to
.
Since the equation is equal to zero, at least one of the expressions on the left side of the equation must be equal to zero. Therefore, set and solve to get
.
Finally, set to get .
The solution to the original equation is
.
Example Question #21 : Finding Roots
Solve for :
To solve the quadratic equation, , let's first set the equation equal to zero and then factor the quadratic to
.
Since the equation is equal to zero, at least one of the expressions on the left side of the equation must be equal to zero. Therefore, set and solve to get
.
Finally, set to get
.
The solution to the original equation is
.
Example Question #22 : Solving Quadratic Equations
Solve for :
In order to solve the equation , first set the expression equal to zero:
Then factor:
Since the equation is equal to zero, at least one of the expressions on the lefthand side of the equation must therefore be equal to zero. Starting with the first expression:
Finally, set the second expression to zero and solve for :
The solution to the equation is
.
Example Question #3 : Finding Zeros Of A Polynomial
Find the roots of .
If we recognize this as an expression with form , with and , we can solve this equation by factoring:
and
and
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