SAT Math : SAT Mathematics

Study concepts, example questions & explanations for SAT Math

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Example Questions

Example Question #892 : Algebra

Find the extraneous solution for 

Possible Answers:

There are no extraneous solutions

Correct answer:

Explanation:

 

Now lets plug these values into our original equation.

 

For 

So this isn't an extraneous solution.

 

For 

Since  is an extraneous solution.

 

 

 

Example Question #1 : Distributive Property

Possible Answers:

Correct answer:

Explanation:

Example Question #2 : Distributive Property

If , what is the value of ?

Possible Answers:

Correct answer:

Explanation:

Remember that (a – b )(a b ) = a 2 – b 2.

We can therefore rewrite (3x – 4)(3x + 4) = 2 as (3x )– (4)2 = 2.

Simplify to find 9x– 16 = 2.

Adding 16 to each side gives us 9x2 = 18.

Example Question #3 : Distributive Property

If  and , then which of the following is equivalent to ?

Possible Answers:

Correct answer:

Explanation:

We are asked to find the difference between g(h(x)) and h(g(x)), where g(x) = 2x2 – 2 and h(x) = x + 4. Let's find expressions for both.

g(h(x)) = g(x + 4) = 2(x + 4)2 – 2

g(h(x)) = 2(x + 4)(x + 4) – 2

In order to find (x+4)(x+4) we can use the FOIL method.

(x + 4)(x + 4) = x2 + 4x + 4x + 16

g(h(x)) = 2(x2 + 4x + 4x + 16) – 2

g(h(x)) = 2(x2 + 8x + 16) – 2

Distribute and simplify.

g(h(x)) = 2x2 + 16x + 32 – 2

g(h(x)) = 2x2 + 16x + 30

Now, we need to find h(g(x)).

h(g(x)) = h(2x2 – 2) = 2x2 – 2 + 4

h(g(x)) = 2x2 + 2

Finally, we can find g(h(x)) – h(g(x)).

g(h(x)) – h(g(x)) = 2x2 + 16x + 30 – (2x2 + 2)

= 2x2 + 16x + 30 – 2x2 – 2

= 16x + 28

The answer is 16x + 28.

Example Question #893 : Algebra

The sum of two numbers is . The product of the same two numbers is . If the two numbers are each increased by one, the new product is . Find  in terms of .

Possible Answers:

Correct answer:

Explanation:

Let the two numbers be x and y.

xy s

xyp

(x + 1)(y + 1) = q

Expand the last equation:

xyxy + 1 = q

Note that both of the first two equations can be substituted into this new equation:

ps + 1 = q

Solve this equation for q – p by subtracting p from both sides:

s + 1 = q – p

Example Question #4 : Distributive Property

Expand the expression:

\dpi{100} \small (x^{3}-4x)(6 + 12x^{2})

Possible Answers:

\dpi{100} \small 6x^{3} + 12x^{2}-24x-48

\dpi{100} \small 22x^{2}

\dpi{100} \small 42x^{3}+12x^{5}-24x

\dpi{100} \small 12x^{5}-42x^{3}-24x

\dpi{100} \small 6x^{3} + 12x^{5}-24x-48x^{3}

Correct answer:

\dpi{100} \small 12x^{5}-42x^{3}-24x

Explanation:

When using FOIL, multiply the first, outside, inside, then last expressions; then combine like terms.

\dpi{100} \small (x^{3}-4x)(6 + 12x^{2})

\dpi{100} \small 6x^{3}+12x^{5}-24x-48x^{3}

\dpi{100} \small -42x^{3}+12x^{5}-24x

\dpi{100} \small 12x^{5}-42x^{3}-24x

Example Question #5 : Distributive Property

Expand the following expression:

(4x+2)(x^2-2)

Possible Answers:

x^3+2x^2-8x-4

4x^3+2x^2-8x-4

4x^3+2x^2+8x+4

4x^3+4x-4

4x^3-4

Correct answer:

4x^3+2x^2-8x-4

Explanation:

(4x+2)(x^2-2)=(4x\times x^2)+(4x\times -2)+(2\times x^2) +(2\times -2)

Which becomes

4x^3-8x+2x^2-4

Or, written better

4x^3+2x^2-8x-4

Example Question #6 : Distributive Property

Which of the following is equal to the expression ?

Possible Answers:

Correct answer:

Explanation:

Multiply using FOIL:

First = 3x(2x) = 6x2

Outter = 3x(4) = 12x

Inner = -1(2x) = -2x

Last = -1(4) = -4

Combine and simplify:

6x2 + 12x - 2x - 4 = 6x2 +10x - 4

Example Question #7 : Distributive Property

Simplify the expression.

Possible Answers:

None of the other answers


Correct answer:

Explanation:

Solve by applying FOIL:

First: 2x2 * 2y = 4x2y

Outer: 2x2 * a = 2ax2

Inner: –3x * 2y = –6xy

Last: –3x * a = –3ax

Add them together: 4x2y + 2ax2 – 6xy – 3ax

There are no common terms, so we are done.

Example Question #8 : Distributive Property

Given the equation above, what is the value of ?

Possible Answers:

Correct answer:

Explanation:

Use FOIL to expand the left side of the equation.

From this equation, we can solve for , , and .

Plug these values into to solve.

 

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