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SAT II Math I : Single-Variable Algebra

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

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All SAT II Math I Resources

6 Diagnostic Tests 113 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #63 : Solving Equations

Solve for .

-8x^2=-144

Possible Answers:

$\pm2\sqrt{3}$

$\pm18$

$3\sqrt{2}$

$2\sqrt{3}$

$\pm3\sqrt{2}$

Correct answer:

$\pm3\sqrt{2}$

Explanation:

-8x^2=-144 Divide on both sides.

x^2=18 Take the square root on both sides. Remember to account for a negative square root. Two negatives multiplied is a positive number.

$x=\pm \sqrt{18}=\pm\sqrt{9}*\sqrt{2}=\pm3\sqrt{2}$

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Example Question #71 : Single Variable Algebra

Solve for .

$\frac{x-8}{8}=\frac{4}{x+6}$

Possible Answers:

10, -8

$\pm10$

$\pm8$

Correct answer:

10, -8

Explanation:

$\frac{x-8}{8}=\frac{4}{x+6}$ Cross-multiply.

(x-8)(x+6)=4*8 Foil out the terms and simplify.

x^2-2x-48=32 Subtract on both sides.

x^2-2x-80=0 We have a quadratic equation. We need to find two terms that multiply to and aso add to .

(x-10)(x+8) Set them individualy equal to zero.

x-10=0 Add to both sides. x=10

x+8=0 Subtract on both sides. x=-8

We should still check the answers.

$\frac{10-8}{8}=\frac{4}{10+6}$ With simplifications, $\frac{2}{8}=\frac{4}{16}=\frac{1}{4}$ . is good.

$\frac{-8-8}{8}=\frac{4}{-8+6}$ With simplifications, $\frac{-16}{8}=\frac{4}{-2}=-2$ . is good.

Answers are 10, -8 .

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Example Question #71 : Single Variable Algebra

Solve for .

4x-15=145

Possible Answers:

160

240

Correct answer:

Explanation:

4x-15=145 Add on both sides.

4x=160 Divide on both sides.

x=40

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Example Question #66 : Solving Equations

Solve for .

3x-83=-11

Possible Answers:

Correct answer:

Explanation:

3x-83=-11 Add on both sides.

3x=72 Divide on both sides.

x=24

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Example Question #67 : Solving Equations

Solve for .

$4\left | x\right |+93=145$

Possible Answers:

$\pm13$

$\pm52$

$\pm12$

$\pm22$

Correct answer:

$\pm13$

Explanation:

$4\left | x\right |+93=145$ Subtract on both sides.

$4\left | x\right |=52$ Divide on both sides.

$\left | x\right |=13$ Since it's absolute value, we need to accept both positive and negative answers.

$x=\pm13$

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Example Question #68 : Solving Equations

Solve for .

$\frac{\sqrt{x}-6}{3}=8$

Possible Answers:

364

900

256

Correct answer:

900

Explanation:

$\frac{\sqrt{x}-6}{3}=8$ Multiply on both sides.

$\sqrt{x}-6=24$ Add on both sides.

$\sqrt{x}=30$ Square both sides to get rid of the radical.

x=30^2=900

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Example Question #69 : Solving Equations

Solve for .

18(x+6)=4x

Possible Answers:

$-\frac{54}{11}$

$-\frac{27}{7}$

$-\frac{54}{7}$

$\frac{27}{7}$

$\frac{54}{11}$

Correct answer:

$-\frac{54}{7}$

Explanation:

18(x+6)=4x Distribute the to each term in the parentheses.

18x+108=4x Subtract 18x on both sides.

108=-14x Divide on both sides.

$x=\frac{-108}{14}=-\frac{54}{7}$

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Example Question #70 : Solving Equations

Solve for .

(x+5)^2=289

Possible Answers:

$\pm22$

$\pm12$

-22, 12

Correct answer:

-22, 12

Explanation:

(x+5)^2=289 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.

x+5=17 Subtract on both sides.

x=12

-(x+5)=17 Divide on both sides.

x+5=-17 Subtract on both sides.

x=-22

Answers are -22, 12 .

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Example Question #73 : Single Variable Algebra

Solve the following equation for when x=5 :

y=5x-25

Possible Answers:

y=5

y=35

y=30

y=6

y=0

Correct answer:

y=0

Explanation:

The first step will be to plug our given variable into the equation to get

$y=5\left ( 5\right )-25$ .

Then you do the multiplication first so it is now,

$\\(5*5=25) \\y=25-25$ .

Finally, subtract from to get y=0 .

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Example Question #71 : Solving Equations

A cubic polynomial p(x) with rational coefficients whose lead term is $x^{3}$ has 2 and 2 + 3i as two of its zeroes. Which of the following is this polynomial?

Possible Answers:

$p (x)= x^{3}-6x^{2} +21x - 26$

$p(x)= x^{3}+2x^{2} +5x - 26$

$p(x)= x^{3}+2x^{2} -13x - 26$

$p(x)= x^{3}-6x^{2} +5x - 26$

$p(x)= x^{3}-2x^{2} +13x - 26$

Correct answer:

$p (x)= x^{3}-6x^{2} +21x - 26$

Explanation:

A cubic polynomial has three zeroes, if a zero of degree is counted times. Since its lead term is $x^{3}$ , we know that, in factored form,

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

where $b_{1}$ , $b_{2}$ , and $b_{3}$ are its zeroes.

A polynomial with rational coefficients has its imaginary zeroes in conjugate pairs. Since p(x) is such a polynomial, then, since 2 + 3i is one of its zeroes, so is its complex conjugate, 2 -3i . It has one other known zero, 2.

Therefore, we can set $b_{1} = 2 + 3i$ , $b_{2} = 2 -3i$ , $b_{3} = 2$ in the factored form of p(x) , and

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

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:

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

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

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

$= x^{2} - 4x+4 +9$

$= x^{2} - 4x+13$

Thus,

$p(x) = (x^{2} - 4x+13)(x-2)$

Distributing:

$= x^{2} (x-2)- 4x(x-2)+13 (x-2)$

$= x^{3}-2x^{2} - 4x^{2}+8x+13x - 26$

$= x^{3}-6x^{2} +21x - 26$

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SAT II Math I : Single-Variable Algebra

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

All SAT II Math I Resources

Example Questions

Example Question #63 : Solving Equations

Example Question #71 : Single Variable Algebra

Example Question #71 : Single Variable Algebra

Example Question #66 : Solving Equations

Example Question #67 : Solving Equations

Example Question #68 : Solving Equations

Example Question #69 : Solving Equations

Example Question #70 : Solving Equations

Example Question #73 : Single Variable Algebra

Example Question #71 : Solving Equations

All SAT II Math I Resources

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