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SAT II Math II : SAT Subject Test in Math II

Study concepts, example questions & explanations for SAT II Math II

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2 Diagnostic Tests 130 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #21 : Factoring And Finding Roots

A polynomial p(x) of degree 4 has as its lead term $x^{4}$ and has rational coefficients. One of its zeroes is 3+ 2i ; this zero has multiplicity two.

Which of the following is this polynomial?

Possible Answers:

$x^{4} - 12x^{3} + 62x^{2} - 156x +169$

$x^{4} +12x^{3} + 62x^{2}+ 156x +169$

Cannot be determined

$x^{4} -12x^{3} + 46x^{2}- 60x +25$

$x^{4} +12x^{3} + 46x^{2}+ 60x +25$

Correct answer:

$x^{4} - 12x^{3} + 62x^{2} - 156x +169$

Explanation:

A fourth-degree, or quartic, polynomial has four zeroes, if a zero of multiplicity is counted times. Since its lead term is $x^{4}$ , we know by the Factor Theorem that

$p(x) = (x-b_{1}) (x-b_{2}) (x-b_{3})(x-b_{4})$

where the $b_{n}$ terms are the four zeroes.

A polynomial with rational coefficients has its imaginary zeroes in conjugate pairs. Since p(x) is such a polynomial, then, since 3+ 2i is a zero of multiplicity 2, so is its complex conjugate 3- 2i . We can set $b_{1} = b_{3} = 4-i$ and $b_{2} = b_{4} = 4+i$ , and

p(x) = [x-(3+ 2i)] [x-(3- 2i)] [x-(3+ 2i)] [x-(3- 2i)]

$p(x) =\left \{[x-(3+ 2i)] [x-(3- 2i)] \right \}^{2}$

We can rewrite this as

$p(x) = \left \{ [x- 3- 2i][x- 3+ 2i ] \right \} ^{2}$

$p(x) =\left \{ [(x- 3)-2i ][(x- 3)+2i ] \right \} ^{2}$

Multiply these factors using the difference of squares pattern, then the square of a binomial pattern:

[(x- 3)-2i ][(x- 3)+2i ]

$= (x- 3)^{2}- (2i )^{2}$

$= (x^{2} - 2 \cdot x \cdot 3 +3^{2})- (2^{2} i^{2} )$

$= x ^{2} -6x+9 -(4 i^{2})$

$= x ^{2} -6x+9 -(-4 )$

$= x ^{2} -6x+13$

Therefore,

$p(x) =( x ^{2} -6x+13 )^{2}$

Multiplying:

$x ^{2} -6x+13$

$\underline{x ^{2} -6x+13}$

$13 x^{2} -78x +169$

$-6x^{3} +36 x ^{2} -78x$

$\underline{x^{4} -6x^{3} +13x^{2}\textup{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }}$

$x^{4} - 12x^{3} + 62x^{2} - 156x +169$

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Example Question #21 : Factoring And Finding Roots

Find the roots to: y=3x^2-1

Possible Answers:

$\pm \frac{1}{3}$

$\pm \sqrt{3}$

$\pm \frac{\sqrt{3}}{2}$

$\pm \frac{1}{9}$

$\pm \frac{\sqrt3}{3}$

Correct answer:

$\pm \frac{\sqrt3}{3}$

Explanation:

This equation cannot be factored. Since this is a parabola, we can use the quadratic equation to find the roots.

y=ax^2+bx+c

$x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substitute the coefficients of a,b,c .

$x = \frac{-0\pm \sqrt{0^2-4(3)(-1)}}{2(3)} =\pm \frac{\sqrt{12}}{6} = \pm\frac{2\sqrt3}{6}$

The answers are: $\pm \frac{\sqrt3}{3}$

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Example Question #541 : Sat Subject Test In Math Ii

Where do the graphs of these two equations intersect?

Equation 1: y = 6x^2 - 3

Equation 2: y = 5x - 2

Possible Answers:

$(1,3)\ \textup{and}\ (-\frac{1}{6},-\frac{17}{6})$

$(1,0)\ \textup{and}\ (-\frac{1}{6},0)$

$(\frac{1}{\sqrt{2}},0)\ \textup{and}\ (\frac{2}{5},0)$

$(1,3)\ \textup{and}\ (-1,3)$

$(\frac{1}{6},0)\ \textup{and}\ (-\frac{1}{6},0)$

Correct answer:

$(1,3)\ \textup{and}\ (-\frac{1}{6},-\frac{17}{6})$

Explanation:

One can find the points of intersection of these two functions by setting them equal to one another, essentially substituting in one equation for the -side of the other equation. This will tell us when the (output) will be the same in each equation for a given (input).

6x^2 - 3 = 5x - 2

By simplifying this equation and setting it equal to zero, we can find the two -values that produce the same values in the system of the two equations.

6x^2 - 3 = 5x - 2

Subtract from both sides of the equation, and add to both sides.

(6x^2 - 3) - (5x - 2) = (5x - 2) - (5x - 2)

(6x^2 - 3) - 5x + 2 = (5x - 2) - 5x + 2

6x^2 - 5x - 1 = 0

Factoring this last equation makes it easier to find the -values that will result in zero on the left side of the equation. Set the two parenthetical phrases equal to zero to find two separate -values that satisfy the equation. These values will be the values of the points of intersection between the two equations.

6x^2 - 5x - 1 = 0

We know our factors multiply to , and the six times one factor plus the other is equal to

1*-1=-1 and 6(-1)+1=-5 , so and are our factors.

(6x+1)(x-1) = 0

$6x+1 = 0 \ \textup{and}\ x-1 = 0$

$x = -\frac{1}{6} \ \textup{and}\ x=1$

Substituting these two values into either of the two original equations results in the -values of the points of intersection.

y_1=5(1)-2=3

$y_2=5(-\frac{1}{6})-2=-\frac{17}{6}$

$(1,3), (-\frac{1}{6}, -\frac{17}{6})$ are the points of intersection.

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Example Question #1 : Systems Of Equations

Find the solution:

3x+2y=12

-2x+3y=5

Possible Answers:

x=2, y=3

x=-2, y=-3

x=3, y=2

x=6, y=3

x=3, y=6

Correct answer:

x=2, y=3

Explanation:

To solve this system of equations, we must first eliminate one of the variables. We will begin by eliminating the variables by finding the least common multiple of the variable's coefficients. The least common multiple of 3 and 2 is 6, so we will multiply each equation in the system by the corresponding number, like

2(3x+2y=12)

3(-2x+3y=5) .

By using the distributive property, we will end up with

6x+4y=24

-6x+9y=15

Now, add down each column so that you have

$0x+13y=39 \rightarrow 13y=39$

Then you solve for and determine that y=3 .

But you're not done yet! To find , you have to plug your answer for back into one of the original equations:

3x+2(3)=12

Solve, and you will find that x=2 .

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Example Question #1 : Systems Of Equations

$\frac{4}{a} + \frac{7}{b} = 18$

$\frac{3}{a} - \frac{5}{b} = 34$

Evaluate

Possible Answers:

The system has no solution.

a = 2

a = 8

$a = \frac{1}{8}$

$a = \frac{1}{2}$

Correct answer:

$a = \frac{1}{8}$

Explanation:

Rewrite the two equations by setting $x = \frac{1}{a}$ and $y = \frac{1}{b}$ and substituting:

$\frac{4}{a} + \frac{7}{b} = 18$

$4 \cdot \frac{1}{a} + 7 \cdot \frac{1 }{b} = 18$

4 x+ 7 y = 18

$\frac{3}{a} - \frac{5}{b} = 34$

$3 \cdot \frac{1}{a} - 5 \cdot \frac{1}{b} = 34$

3x-5y = 34

The result is a two-by-two linear system in terms of and :

4x+7y = 18

3x-5y = 34

This can be solved, among other ways, using Gaussian elimination on an augmented matrix:

$\begin{bmatrix} 4& 7\\ 3& -5 \end{matrix} \begin{vmatrix} 18 \\ 34 \\ \end{bmatrix}$

Perform row operations until the left two columns show identity matrix $\begin{bmatrix} 1 & 0\\ 0 &1 \end{bmatrix}$ . One possible sequence:

$R1-R2 \rightarrow R1$

$\begin{bmatrix} 4 - 3 & 7- (-5 )\\ 3& -5 \end{matrix} \begin{vmatrix} 18 - 34 \\ 34 \\ \end{bmatrix}$

$\begin{bmatrix} 1 & 12 \\ 3& -5 \end{matrix} \begin{vmatrix} -16 \\ 34 \\ \end{bmatrix}$

$R2 - 3R1 \rightarrow R2$

$\begin{bmatrix} 1 & 12 \\ 3 - 3(1)& -5 - 3(12) \end{matrix} \begin{vmatrix} -16 \\ 34 - 3(-16) \\ \end{bmatrix}$

$\begin{bmatrix} 1 & 12 \\ 0& -41\end{matrix} \begin{vmatrix} -16 \\ 82\\ \end{bmatrix}$

$R2 \div (-41 )\rightarrow R2$

$\begin{bmatrix} 1 & 12 \\ 0\div (-41 )& -41\div (-41 )\end{matrix} \begin{vmatrix} -16 \\ 82\div (-41 )\\ \end{bmatrix}$

$\begin{bmatrix} 1 & 12 \\ 0& 1 \end{matrix} \begin{vmatrix} -16 \\-2\\ \end{bmatrix}$

$R1 - 12 R2 \rightarrow R1$

$\begin{bmatrix} 1 - 12(0) & 12 - 12(1) \\ 0& 1 \end{matrix} \begin{vmatrix} -16 - 12(-2) \\ 2\\ \end{bmatrix}$

$\begin{bmatrix} 1 & 0 \\ 0& 1 \end{matrix} \begin{vmatrix} 8 \\ 2\\ \end{bmatrix}$

x= 8 and y = 2 .

In the former equation, substitute $\frac{1}{a}$ back for :

$\frac{1}{a} = 8$ .

Taking the reciprocal of both sides, we get

$a = \frac{1}{8}$ .

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Example Question #1 : Systems Of Equations

$4 \sqrt [3] {a} + 7 \sqrt [3]{b} = 18$

$3 \sqrt [3]{a} -5 \sqrt [3]{b} = 34$

Evaluate .

Possible Answers:

b = 16

b = 2

b= 8

b= 512

The system has no solution.

Correct answer:

b= 8

Explanation:

Rewrite the two equations by setting $x = \sqrt [3] {a}$ and $y = \sqrt [3] {b}$ and substituting:

$4 \sqrt [3] {a} + 7 \sqrt [3]{b} = 18$

4 x+ 7 y = 18

$3 \sqrt [3]{a} -5 \sqrt [3]{b} = 34$

3x-5y = 34

The result is a two-by-two linear system in terms of and :

4x+7y = 18

3x-5y = 34

This can be solved, among other ways, using Gaussian elimination on an augmented matrix:

$\begin{bmatrix} 4& 7\\ 3& -5 \end{matrix} \begin{vmatrix} 18 \\ 34 \\ \end{bmatrix}$

Perform row operations until the left two columns show identity matrix $\begin{bmatrix} 1 & 0\\ 0 &1 \end{bmatrix}$ . One possible sequence:

$R1-R2 \rightarrow R1$

$\begin{bmatrix} 4 - 3 & 7- (-5 )\\ 3& -5 \end{matrix} \begin{vmatrix} 18 - 34 \\ 34 \\ \end{bmatrix}$

$\begin{bmatrix} 1 & 12 \\ 3& -5 \end{matrix} \begin{vmatrix} -16 \\ 34 \\ \end{bmatrix}$

$R2 - 3R1 \rightarrow R2$

$\begin{bmatrix} 1 & 12 \\ 3 - 3(1)& -5 - 3(12) \end{matrix} \begin{vmatrix} -16 \\ 34 - 3(-16) \\ \end{bmatrix}$

$\begin{bmatrix} 1 & 12 \\ 0& -41\end{matrix} \begin{vmatrix} -16 \\ 82\\ \end{bmatrix}$

$R2 \div (-41 )\rightarrow R2$

$\begin{bmatrix} 1 & 12 \\ 0\div (-41 )& -41\div (-41 )\end{matrix} \begin{vmatrix} -16 \\ 82\div (-41 )\\ \end{bmatrix}$

$\begin{bmatrix} 1 & 12 \\ 0& 1 \end{matrix} \begin{vmatrix} -16 \\-2\\ \end{bmatrix}$

$R1 - 12 R2 \rightarrow R1$

$\begin{bmatrix} 1 - 12(0) & 12 - 12(1) \\ 0& 1 \end{matrix} \begin{vmatrix} -16 - 12(-2) \\ 2\\ \end{bmatrix}$

$\begin{bmatrix} 1 & 0 \\ 0& 1 \end{matrix} \begin{vmatrix} 8 \\ 2\\ \end{bmatrix}$

x= 8 and y = 2 .

Substituting back in the second equation and solving for by cubing both sides:

$\sqrt [3] {b}= 2$

$(\sqrt [3] {b})^{3}= 2^{3}$

b= 8

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Example Question #93 : Single Variable Algebra

$4 \sqrt{ a} + 3 \sqrt{b} = 27$

$5 \sqrt{ a} + 2 \sqrt{b} = 25$

Evaluate .

Possible Answers:

a=9

The system has no solution.

a= 3

a= 25

a = 5

Correct answer:

a=9

Explanation:

Rewrite the two equations by setting $x = \sqrt {a}$ and $y = \sqrt {b}$ and substituting:

$4 \sqrt{ a} + 3 \sqrt{b} = 27$

4x+3y = 27

$5 \sqrt{ a} + 2 \sqrt{b} = 25$

5x+2y = 25

In terms of and , this is a two-by-two linear system:

4x+3y = 27

5x+2y = 25

Rewrite this as an augmented matrix as follows:

$\begin{bmatrix} 4& 3\\ 5& 2 \end{matrix} \begin{vmatrix} 27\\ 25 \end{bmatrix}$

Perform the following row operations to make the left two columns the identity matrix $\begin{bmatrix}1 & 0\\ 0 &1 \end{bmatrix}$ :

$R1 \leftrightarrow R2$

$\begin{bmatrix} 5& 2 \\ 4& 3 \end{matrix} \begin{vmatrix} 25 \\ 27 \end{bmatrix}$

$R1-R2 \rightarrow R1$

$\begin{bmatrix} 5 -4& 2 -3 \\ 4& 3 \end{matrix} \begin{vmatrix} 25 -27 \\ 27 \end{bmatrix}$

$\begin{bmatrix} 1& -1 \\ 4& 3 \end{matrix} \begin{vmatrix} -2 \\ 27 \end{bmatrix}$

$R2 - 4R1 \rightarrow R2$

$\begin{bmatrix} 1& -1 \\ 4 - 4(1)& 3- 4(-1) \end{matrix} \begin{vmatrix} -2 \\ 27-4(-2) \end{bmatrix}$

$\begin{bmatrix} 1& -1 \\ 0& 7 \end{matrix} \begin{vmatrix} -2 \\ 35 \end{bmatrix}$

$R2 \div 7 \rightarrow R2$

$\begin{bmatrix} 1& -1 \\ 0 \div 7 & 7 \div 7 \end{matrix} \begin{vmatrix} -2 \\ 35 \div 7 \end{bmatrix}$

$\begin{bmatrix} 1& -1 \\ 0 & 1 \end{matrix} \begin{vmatrix} -2 \\ 5 \end{bmatrix}$

$R1 +R2 \rightarrow R1$

$\begin{bmatrix} 1+ 0& -1+1 \\ 0 & 1 \end{matrix} \begin{vmatrix} -2+5 \\ 5 \end{bmatrix}$

$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{matrix} \begin{vmatrix} 3 \\ 5 \end{bmatrix}$

x = 3 and y = 5 .

Replacing with $\sqrt {a}$ , then squaring both sides:

$\sqrt {a} = 3$

$(\sqrt {a}) ^{2}= 3^{2}$

a=9

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Example Question #6 : Systems Of Equations

Solve for : $\begin{bmatrix} 2x-8y=3\\x+5y = 1 \end{bmatrix}$

Possible Answers:

$\frac{7}{32}$

$-\frac{5}{32}$

$\frac{23}{18}$

$-\frac{1}{18}$

$\frac{13}{6}$

Correct answer:

$\frac{23}{18}$

Explanation:

To solve for , we will need to eliminate the variable. To do so, we can choose either the substitution or the elimination method.

Let's choose the substitution method. Rewrite the second equation so that is isolated.

Subtract on both sides.

x+5y -x= 1-x

5y = -x+1

Divide by five on both sides.

$\frac{5y}{5} = \frac{-x+1}{5}$

$y=- \frac{1}{5}x+\frac{1}{5}$

Substitute this back to the first equation.

$2x-8(- \frac{1}{5}x+\frac{1}{5})=3$

Distribute the terms.

$2x+\frac{8}{5}x-\frac{8}{5} = 3$

Multiply both sides by 5 to eliminate the fractions.

$5(2x+\frac{8}{5}x-\frac{8}{5}) = 3\cdot 5$

10x+8x-8 = 15

Add 8 on both sides and combine like-terms on the left.

18x-8+8 = 15+8

18x = 23

Divide by 18 on both sides.

$\frac{18x}{18} = \frac{23}{18}$

The answer is: $\frac{23}{18}$

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Example Question #2 : Systems Of Equations

Solve for : $\begin{bmatrix} 3x+2y = 1\\ x-y=-1 \end{bmatrix}$

Possible Answers:

$\frac{4}{3}$

$-\frac{1}{5}$

$\frac{1}{2}$

Correct answer:

$-\frac{1}{5}$

Explanation:

In order to solve for the x-variable, we will need to use the elimination method to eliminate the y-variable.

Multiply the second equation by two.

$\begin{bmatrix} 3x+2y = 1\\ x-y=-1 \end{bmatrix} \rightarrow\begin{bmatrix} 3x+2y = 1\\ 2x-2y=-2 \end{bmatrix}$

Add both equations.

5x=-1

Divide by five on both sides.

$\frac{5x}{5}=\frac{-1}{5}$

The answer is: $-\frac{1}{5}$

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SAT II Math II : SAT Subject Test in Math II

Study concepts, example questions & explanations for SAT II Math II

All SAT II Math II Resources

Example Questions

Example Question #21 : Factoring And Finding Roots

Example Question #21 : Factoring And Finding Roots

Example Question #541 : Sat Subject Test In Math Ii

Example Question #1 : Systems Of Equations

Example Question #1 : Systems Of Equations

Example Question #1 : Systems Of Equations

Example Question #93 : Single Variable Algebra

Example Question #6 : Systems Of Equations

Example Question #2 : Systems Of Equations

All SAT II Math II Resources

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