All SAT II Math I Resources
Example Questions
Example Question #66 : Solving Equations
Solve for .
Add
on both sides.
Divide
on both sides.
Example Question #231 : Sat Subject Test In Math I
Solve for .
Subtract
on both sides.
Divide
on both sides.
Since it's absolute value, we need to accept both positive and negative answers.
Example Question #61 : Solving Equations
Solve for .
Multiply
on both sides.
Add
on both sides.
Square both sides to get rid of the radical.
Example Question #231 : Sat Subject Test In Math I
Solve for .
Distribute the
to each term in the parentheses.
Subtract
on both sides.
Divide
on both sides.
Example Question #72 : Single Variable Algebra
Solve for .
Take the square root on both sides. When you do that, you also need to consider both positive and negative values. Remember, two negatives multiplied create a positive number.
Subtract
on both sides.
Divide
on both sides.
Subtract
on both sides.
Answers are .
Example Question #71 : Single Variable Algebra
Solve the following equation for when :
The first step will be to plug our given variable into the equation to get
.
Then you do the multiplication first so it is now,
.
Finally, subtract from
to get
.
Example Question #71 : Solving Equations
A cubic polynomial with rational coefficients whose lead term is
has 2 and
as two of its zeroes. Which of the following is this polynomial?
A cubic polynomial has three zeroes, if a zero of degree is counted
times. Since its lead term is
, we know that, in factored form,
,
where ,
, and
are its zeroes.
A polynomial with rational coefficients has its imaginary zeroes in conjugate pairs. Since is such a polynomial, then, since
is one of its zeroes, so is its complex conjugate,
. It has one other known zero, 2.
Therefore, we can set ,
,
in the factored form of
, and
To rewrite this, first multiply the first two factors with the help of the difference of squares pattern and the square of a binomial pattern:
Thus,
Distributing:
Example Question #231 : Sat Subject Test In Math I
Which of the following is a factor of the polynomial ?
Call
By the Rational Zeroes Theorem, since has only integer coefficients, any rational solution of
must be a factor of 54 divided by a factor of 1 - positive or negative. 54 has as its factors 1, 2, 3, 6, 9, 18, 27 , 54; 1 has only itself as a factor. Therefore, the rational solutions of
must be chosen from this set:
.
By the Factor Theorem, a polynomial is divisible by
if and only if
- that is, if
is a zero. By the preceding result, we can immediately eliminate
and
- that is,
and
- as factors, since
and
have been eliminated as possible zeroes.
Of the three remaining choices, we can demonstrate that - that is,
- is the factor by evaluating
:
, so
, or, equivalently,
, is a factor of
.
Of the remaining two choices, and
, both can be proved to not be factors by showing that
and
are both nonzero:
, so
is not a factor.
, so
is not a factor.
Example Question #231 : Sat Subject Test In Math I
Each of the following is a factor of the polynomial , except:
Set
By the Rational Zeroes Theorem, a rational number can be a zero of a polynomial if and only if it belongs to the set of numbers that are the quotient of a factor of the constant and a factor of the leading coefficient (positive or negative). Constant 105 has as its factors 1, 3, 5, 7, 15 21, 35, and 105, and lead coefficient 1 has only factor 1, so any rational zeroes must come from the set
By the Factor Theorem, a linear binomial is a factor of
if and only if
is a zero of
. From the set
, only 2 can be eliminated as a zero by the above result, so, of the five choices,
is the one that is eliminated as a factor of
.
The other four can be confirmed to be factors by demonstrating that
,
but this is not necessary.
Example Question #82 : Single Variable Algebra
Define functions and
.
for exactly one value of
on the interval
. Which of the following is true of
?
Define
Then if ,
it follows that
,
or, equivalently,
.
By the Intermediate Value Theorem (IVT), if is a continuous function, and
and
are of unlike sign, then
for some
.
Since polynomial and exponential function
are continuous everywhere, so is
, so the IVT applies here.
Evaluate for each of the following values:
Only in the case of does it hold that
assumes a different sign at both endpoints -
. By the IVT,
, and
, for some
.
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All SAT II Math I Resources
