How to add exponential variables

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ISEE Upper Level Quantitative Reasoning › How to add exponential variables

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1

Simplify:

$\frac{(x^{2}+y^{2})^{2} - (x^{2} - y^{2})^{2}}{x^{4}y^{4}}$

$\frac{4}{x^{2}y^{2}}$

CORRECT

$\frac{4}{x^{4}y^{4}}$

0

$\frac{4}{xy}$

0

0

$\frac{4}{x^{3}y^{3}}$

0

Explanation

$\frac{(x^{2}+y^{2})^{2} - (x^{2} - y^{2})^{2}}{x^{4}y^{4}}$

$=\frac{(x^{4}+2x^{2}y^{2}+y^{4}) - (x^{4}-2x^{2}y^{2}+y^{4}) }{x^{4}y^{4}}$

$=\frac{(x^{4}-x^{4}+2x^{2}y^{2}+2x^{2}y^{2}+y^{4}-y^{4}) }{x^{4}y^{4}}$

$=\frac{4x^{2}y^{2}}{x^{4}y^{4}}$

$=\frac{4}{x^{2}y^{2}}$

2

Rewrite the polynomial in standard form:

$9x^{4} - 3x ^{2} + 8x^{5} -2x + 7 - 4x^{3}$

$8x^{5} + 9x^{4}- 4x^{3} - 3x ^{2} -2x + 7$

CORRECT

$7-2x- 3x ^{2}- 4x^{3}+ 9x^{4} + 8x^{5}$

0

$9x^{4} + 8x^{5} + 7- 4x^{3} - 3x ^{2} -2x$

0

$-2x - 3x ^{2}- 4x^{3}+ 7+ 8x^{5} + 9x^{4}$

0

$9x^{4} + 8x^{5} + 7-2x - 3x ^{2} - 4x^{3}$

0

Explanation

The degree of a term of a polynomial with one variable is the exponent of that variable. The terms of a polynomial in standard form are written in descending order of degree. Therefore, we rearrange the terms by their exponent, from 5 down to 0, noting that we can rewrite the and constant terms with exponents 1 and 0, respectively:

$9x^{4} - 3x ^{2} + 8x^{5} -2x + 7 - 4x^{3}$

$= 9x^{4} - 3x ^{2} + 8x^{5} -2x ^{1} + 7x ^{0} - 4x^{3}$

$=8x^{5} + 9x^{4}- 4x^{3} - 3x ^{2} -2x ^{1}+ 7x^{0}$

$=8x^{5} + 9x^{4}- 4x^{3} - 3x ^{2} -2x+ 7$

3

Which is the greater quantity?

(a) $x^{2} + 3x$

(b)

It is impossible to tell from the information given

CORRECT

(a) and (b) are equal

0

(a) is greater

0

(b) is greater

0

Explanation

We give at least one positive value of for which (a) is greater and at least one positive value of for which (b) is greater.

Case 1:

(a) $x^{2} + 3x = 2^{2} + 3 \cdot 2 = 4 + 6 = 10$

(b) $4x = 4 \cdot 2 = 8$

Case 2: $x = \frac{1}{2}$

(a) $x^{2} + 3x = \left ( \frac{1}{2} \right ) ^{2} + 3 \cdot \frac{1}{2} = \frac{1}{4} + \frac{3}{2} = \frac{7}{4}= 1 \frac{3}{4}$

(b) $4x = 4 \cdot \frac{1}{2} = 2$

Therefore, either (a) or (b) can be greater.

4

Which is the greater quantity?

(a) $(x + y)^{2}$

(b) $x^{2}+ y^{2}$

(b) is greater.

CORRECT

(a) and (b) are equal.

0

(a) is greater.

0

It is impossible to tell from the information given.

0

Explanation

$(x + y)^{2} = x^{2}+2xy + y^{2}$

Since and have different signs,

, and, subsequently,

Therefore,

$(x + y)^{2} = x^{2}+2xy + y^{2} < x^{2} + y^{2}$

This makes (b) the greater quantity.

5

Add:

CORRECT

0

0

0

Explanation

6

Define

What is $g(x ^{2} + 3x)$ ?

$2 x ^{2} + 6x - 5$

CORRECT

$2 x ^{2} + 6x - 10$

0

$2 x ^{2} + 3x - 5$

0

$x ^{2} + 6x - 5$

0

Explanation

Substitute $x ^{2} + 3x$ for in the definition:

$g(x ^{2} + 3x ) = 2 \left ( x ^{2} + 3x \right ) - 5$

$g(x ^{2} + 3x ) = 2 \left ( x ^{2}\right ) + 2 \left ( 3x \right ) - 5$

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

7

Assume that and are not both zero. Which is the greater quantity?

(a) $\frac{(x+y)^{2} + (x - y)^{2}}{x^{2}+y^{2}}$

(b)

It is impossible to tell from the information given.

CORRECT

(a) is greater.

0

(b) is greater.

0

(a) and (b) are equal.

0

Explanation

Simplify the expression in (a):

$\frac{(x+y)^{2} + (x - y)^{2}}{x^{2}+y^{2}}$

$=\frac{(x^{2}+2xy + y^{2})+ (x^{2}-2xy + y^{2})}{x^{2}+y^{2}}$

$=\frac{ x^{2}+x^{2} +2xy -2xy+ y^{2}+ y^{2}}{x^{2}+y^{2}}$

$=\frac{ 2x^{2} +2 y^{2}}{x^{2}+y^{2}} =\frac{ 2(x^{2}+y^{2})}{x^{2}+y^{2}} = 2$

Therefore, whether (a) or (b) is greater depends on the values of and , neither of which are known.

8

Assume all variables to be nonzero.

Simplify: $\left (12x^{5}y ^{4}z^{3} \right )^{0} + \left (3x^{5}y ^{4}z^{3} \right )^{0}$

None of the answer choices are correct.

CORRECT

$15x^{5}y ^{4}z^{3}$

0

$36x^{5}y ^{4}z^{3}$

0

$15x^{10}y ^{8}z^{6}$

0

$36x^{10}y ^{8}z^{6}$

0

Explanation

Any nonzero expression raised to the power of 0 is equal to 1. Therefore,

$\left (12x^{5}y ^{4}z^{3} \right )^{0} + \left (3x^{5}y ^{4}z^{3} \right )^{0} = 1 + 1 = 2$ .

None of the given expressions are correct.

9

Simplify:

CORRECT

0

0

0

0

Explanation

Expand each term by using FOIL:

Rearrange to group like-terms together.

Simplify by combining like-terms.

10

Assume that $x,y \neq 0$ . Simplify:

$\frac{(x^{2}+y^{2})^{2} +(x^{2} - y^{2})^{2}}{x^{4}y^{4}}$

$\frac{2x^{4}+2y^{4} }{x^{4}y^{4}}$

CORRECT

$\frac{2x^{4}-2y^{4} }{x^{4}y^{4}}$

0

$\frac{4}{x^{2}y^{2}}$

0

$2x^{4}-2y^{4}$

0

$2x^{4}+2y^{4}$

0

Explanation

$\frac{(x^{2}+y^{2})^{2} + (x^{2} + y^{2})^{2}}{x^{4}y^{4}}$

$=\frac{(x^{4}+2x^{2}y^{2}+y^{4}) + (x^{4}-2x^{2}y^{2}+y^{4}) }{x^{4}y^{4}}$

$=\frac{x^{4}+x^{4}+2x^{2}y^{2}-2x^{2}y^{2}+y^{4}+y^{4} }{x^{4}y^{4}}$

$=\frac{2x^{4}+2y^{4} }{x^{4}y^{4}}$

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