How to add exponential variables - ISEE Upper Level: Quantitative Reasoning

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Question

Simplify:

$5x^{2} y ^{2}+7 + 6x^{2}y^{2} + 11 xy$

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Answer

Group and combine like terms $5x^{2} y ^{2},6x^{2}y^{2}$ :

$5x^{2} y ^{2}+7 + 6x^{2}y^{2} + 11 xy$

$= 5x^{2} y ^{2}+ 6x^{2}y^{2}+ 11 xy +7$

$=\left ( 5+ 6 \right )x^{2}y^{2}+ 11 xy +7$

$=11x^{2}y^{2}+ 11 xy +7$

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Question

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)

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Answer

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.

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Question

Which is the greater quantity?

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

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

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Answer

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

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.

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Question

Which is the greater quantity?

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

(b)

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Answer

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.

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Question

Assume all variables to be nonzero.

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

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Answer

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.

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Question

Add:

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Answer

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Question

Rewrite the polynomial in standard form:

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

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Answer

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:

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

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

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

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

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Question

Simplify:

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

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Answer

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

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

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

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

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

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Question

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

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

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Answer

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

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

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

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

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Question

If $x+y\neq 0$ , simplify:

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

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Answer

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

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

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

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

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Question

If $x\neq 0$ , simplify:

$\frac{(x+\frac{1}{x})^3+(x-\frac{1}{x})^3}{\frac{2}{x}}$

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Answer

$\frac{(x+\frac{1}{x})^3+(x-\frac{1}{x})^3}{\frac{2}{x}}$

$=\frac{(x^3+3x^2\times \frac{1}{x}+3x\times \frac{1}{x^2}+\frac{1}{x^3})+(x^3-3x^2\times \frac{1}{x}+3x\times \frac{1}{x^2}-\frac{1}{x^3})}{\frac{2}{x}}$

$=(x^3+x^3)+(3x-3x)+(\frac{3}{x}+\frac{3}{x})+(\frac{1}{x^3}-\frac{1}{x^3})$

$=\frac{2x^3+\frac{6}{x}}{\frac{2}{x}}$

$=\frac{\frac{2x^4+6}{x}}{\frac{2}{x}}$

$=\frac{2x^4+6}{2}$

$=\frac{2(x^4+3)}{2}$

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Question

Define

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

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Answer

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

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

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

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

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Question

Simplify:

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Answer

Start by reordering the expression to group like-terms together.

Combine like-terms to simplify.

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Question

Simplify:

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Answer

We can expand the first term using FOIL:

Reorder the expression to group like-terms together.

Simplify by combining like-terms.

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Question

Simplify:

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Answer

Expand each term by using FOIL:

Rearrange to group like-terms together.

Simplify by combining like-terms.

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Question

Simplify:

$5x^{2} y ^{2}+7 + 6x^{2}y^{2} + 11 xy$

Tap to reveal answer

Answer

Group and combine like terms $5x^{2} y ^{2},6x^{2}y^{2}$ :

$5x^{2} y ^{2}+7 + 6x^{2}y^{2} + 11 xy$

$= 5x^{2} y ^{2}+ 6x^{2}y^{2}+ 11 xy +7$

$=\left ( 5+ 6 \right )x^{2}y^{2}+ 11 xy +7$

$=11x^{2}y^{2}+ 11 xy +7$

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Question

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)

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Answer

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.

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Question

Which is the greater quantity?

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

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

Tap to reveal answer

Answer

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

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.

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Question

Which is the greater quantity?

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

(b)

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Answer

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.

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Question

Assume all variables to be nonzero.

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

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Answer

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.

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