Other Squaring / Square Roots - SAT Math

Card 0 of 64

Question

How many integers from 20 to 80, inclusive, are NOT the square of another integer?

Answer

First list all the integers between 20 and 80 that are squares of another integer:

52 = 25

62 = 36

72 = 49

82 = 64

In total, there are 61 integers from 20 to 80, inclusive. 61 – 4 = 57

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Question

The square root of 5184 is:

Answer

The easiest way to narrow down a square root from a list is to look at the last number on the squared number – in this case 4 – and compare it to the last number of the answer.

70 * 70 will equal XXX0

71 * 71 will equal XXX1

72 * 72 will equal XXX4

73 * 73 will equal XXX9

74 * 74 will equal XXX(1)6

Therefore 72 is the answer. Check by multiplying it out.

Compare your answer with the correct one above

Question

If x and y are integers and xy + y2 is even, which of the following statements must be true?

I. 3y is odd

II. y/2 is an integer

III. xy is even

Answer

In order for the original statement to be true, the and $y^{2}$ terms must be either both odd or both even. Looking at each of the statements individually,

I. States that is odd, but only odd values multiplied by 3 are odd. If was an even number, the result would be even. But can be either odd or even, depending on what equals. Thus this statement COULD be true but does not HAVE to be true.

II. States that $\frac{y}{2}$ is an integer, and since only even numbers are cleanly divided by 2 (odd values result in a fraction) this ensures that is even. However, can also be odd, so this is a statement that COULD be true but does not HAVE to be true.

III. For exponents, only the base value determines whether it is even or odd - it does not indicate the value of y at all. Only even numbers raised to any power are even, thus, this ensures that is even. But can be odd as well, so this statement COULD be true but does not HAVE to be true.

An example of two integers that will work violate conditions II and III is and .

$(1)\cdot (3)+(3)^{2}=3+9=12$ , and even number.

$\frac{3}{2}$ is not an integer.

$1^{3}=1$ is not even.

Furthermore, any combination of 2 even integers will make the original statement true, and violate the Statement I.

Compare your answer with the correct one above

Question

Consider the inequality:

$x< x^{5}< x^{4}$

Which of the following could be a value of ?

Answer

Notice how $x^4$ is the greatest value. This often means that $x$ is negative as $(-1)^n=-1$ when $\dpi{100} n$ is odd and $(-1)^n=1$ when $\dpi{100} n$ is even.

Let us examine the first choice, $x=-\frac{3}{4}$

$x^5=-\frac{3^5}{4^5}=-\frac{243}{1024}> -\frac{3}{4}$

This can only be true of a negative value that lies between zero and one.

Compare your answer with the correct one above

Question

If $5^{x}+5^{x}+5^{x}+5^{x}+5^{x}=5^{10}$ , what is the value of x?

Answer

$5(5^{x})=5^{10}$

$5^{1}(5^{x})=5^{10}$

$5^{x+1}=5^{10}$

$x=9$

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Question

Solve each problem and decide which is the best of the choices given.

Solve for .

Answer

To solve for you have to get all the 's on one side and all the numbers on the other side.

Combine like terms, and the result is

.

Divide by on both sides to get

.

Finally take the square root of both sides to get $x=\pm\sqrt{27}$ .

Compare your answer with the correct one above

Question

Give the positive solution of the question.

Answer

This question is testing one's ability to analyze a function algebraically and recognize different but equivalent forms. Identifying properties of functions through analyzing equivalent forms is critical to this concept. Such properties that can be found through analyzing the different forms of a function include finding roots (zeros), extreme values, symmetry, and intercepts.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Recognize the general form of the function.

$\y=a^2x^2-b^2 \y=(ax-b)(ax+b)$

This is known as the difference of squares.

Step 2: Identify what is known.

$\a^2=1\rightarrow a=1 \b^2=16\rightarrow b=4$

Step 3: Substitute the known values into the difference of perfect squares found in step 1.

$\y=(1x-4)(1x+4) \y=(x-4)(x+4)$

Step 4: Answer the question.

To find the positive solution set each binomial equal to zero and solve for x.

$\x-4=0 \x=4$ $\x+4=0 \x=-4$

Therefore the positive solution is four.

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Question

If the following are true, solve for .

$\a=3\sqrt{2} \3a=\sqrt{6x}$

Answer

Given the two equations, substitute the numerical value of into the second equation to solve for .

$\a=3\sqrt{2} \3a=\sqrt{6x}$

Substituting the numerical value for into the equation with is as follows.

$3(3\sqrt{2})=\sqrt{6x}$

From here, distribute the three.

$9\sqrt{2}=\sqrt{6x}$

Now square both side of the equation.

$(9\sqrt{2})^2=(\sqrt{6x})^2$

Remember to square both terms within the parentheses. Also, recall that squaring a square root sign cancels them out.

$\9^2\sqrt{2}^2=6x \81(2)=6x \162=6x$

To solve for divide both sides by six.

$\\frac{162}{6}=\frac{6x}{6} \ \\frac{27\cdot 6}{6}=x \\27=x$

Compare your answer with the correct one above

Question

How many integers from 20 to 80, inclusive, are NOT the square of another integer?

Answer

First list all the integers between 20 and 80 that are squares of another integer:

52 = 25

62 = 36

72 = 49

82 = 64

In total, there are 61 integers from 20 to 80, inclusive. 61 – 4 = 57

Compare your answer with the correct one above

Question

The square root of 5184 is:

Answer

The easiest way to narrow down a square root from a list is to look at the last number on the squared number – in this case 4 – and compare it to the last number of the answer.

70 * 70 will equal XXX0

71 * 71 will equal XXX1

72 * 72 will equal XXX4

73 * 73 will equal XXX9

74 * 74 will equal XXX(1)6

Therefore 72 is the answer. Check by multiplying it out.

Compare your answer with the correct one above

Question

If x and y are integers and xy + y2 is even, which of the following statements must be true?

I. 3y is odd

II. y/2 is an integer

III. xy is even

Answer

In order for the original statement to be true, the and $y^{2}$ terms must be either both odd or both even. Looking at each of the statements individually,

I. States that is odd, but only odd values multiplied by 3 are odd. If was an even number, the result would be even. But can be either odd or even, depending on what equals. Thus this statement COULD be true but does not HAVE to be true.

II. States that $\frac{y}{2}$ is an integer, and since only even numbers are cleanly divided by 2 (odd values result in a fraction) this ensures that is even. However, can also be odd, so this is a statement that COULD be true but does not HAVE to be true.

III. For exponents, only the base value determines whether it is even or odd - it does not indicate the value of y at all. Only even numbers raised to any power are even, thus, this ensures that is even. But can be odd as well, so this statement COULD be true but does not HAVE to be true.

An example of two integers that will work violate conditions II and III is and .

$(1)\cdot (3)+(3)^{2}=3+9=12$ , and even number.

$\frac{3}{2}$ is not an integer.

$1^{3}=1$ is not even.

Furthermore, any combination of 2 even integers will make the original statement true, and violate the Statement I.

Compare your answer with the correct one above

Question

Consider the inequality:

$x< x^{5}< x^{4}$

Which of the following could be a value of ?

Answer

Notice how $x^4$ is the greatest value. This often means that $x$ is negative as $(-1)^n=-1$ when $\dpi{100} n$ is odd and $(-1)^n=1$ when $\dpi{100} n$ is even.

Let us examine the first choice, $x=-\frac{3}{4}$

$x^5=-\frac{3^5}{4^5}=-\frac{243}{1024}> -\frac{3}{4}$

This can only be true of a negative value that lies between zero and one.

Compare your answer with the correct one above

Question

If $5^{x}+5^{x}+5^{x}+5^{x}+5^{x}=5^{10}$ , what is the value of x?

Answer

$5(5^{x})=5^{10}$

$5^{1}(5^{x})=5^{10}$

$5^{x+1}=5^{10}$

$x=9$

Compare your answer with the correct one above

Question

Solve each problem and decide which is the best of the choices given.

Solve for .

Answer

To solve for you have to get all the 's on one side and all the numbers on the other side.

Combine like terms, and the result is

.

Divide by on both sides to get

.

Finally take the square root of both sides to get $x=\pm\sqrt{27}$ .

Compare your answer with the correct one above

Question

Give the positive solution of the question.

Answer

This question is testing one's ability to analyze a function algebraically and recognize different but equivalent forms. Identifying properties of functions through analyzing equivalent forms is critical to this concept. Such properties that can be found through analyzing the different forms of a function include finding roots (zeros), extreme values, symmetry, and intercepts.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Recognize the general form of the function.

$\y=a^2x^2-b^2 \y=(ax-b)(ax+b)$

This is known as the difference of squares.

Step 2: Identify what is known.

$\a^2=1\rightarrow a=1 \b^2=16\rightarrow b=4$

Step 3: Substitute the known values into the difference of perfect squares found in step 1.

$\y=(1x-4)(1x+4) \y=(x-4)(x+4)$

Step 4: Answer the question.

To find the positive solution set each binomial equal to zero and solve for x.

$\x-4=0 \x=4$ $\x+4=0 \x=-4$

Therefore the positive solution is four.

Compare your answer with the correct one above

Question

If the following are true, solve for .

$\a=3\sqrt{2} \3a=\sqrt{6x}$

Answer

Given the two equations, substitute the numerical value of into the second equation to solve for .

$\a=3\sqrt{2} \3a=\sqrt{6x}$

Substituting the numerical value for into the equation with is as follows.

$3(3\sqrt{2})=\sqrt{6x}$

From here, distribute the three.

$9\sqrt{2}=\sqrt{6x}$

Now square both side of the equation.

$(9\sqrt{2})^2=(\sqrt{6x})^2$

Remember to square both terms within the parentheses. Also, recall that squaring a square root sign cancels them out.

$\9^2\sqrt{2}^2=6x \81(2)=6x \162=6x$

To solve for divide both sides by six.

$\\frac{162}{6}=\frac{6x}{6} \ \\frac{27\cdot 6}{6}=x \\27=x$

Compare your answer with the correct one above

Question

How many integers from 20 to 80, inclusive, are NOT the square of another integer?

Answer

First list all the integers between 20 and 80 that are squares of another integer:

52 = 25

62 = 36

72 = 49

82 = 64

In total, there are 61 integers from 20 to 80, inclusive. 61 – 4 = 57

Compare your answer with the correct one above

Question

The square root of 5184 is:

Answer

The easiest way to narrow down a square root from a list is to look at the last number on the squared number – in this case 4 – and compare it to the last number of the answer.

70 * 70 will equal XXX0

71 * 71 will equal XXX1

72 * 72 will equal XXX4

73 * 73 will equal XXX9

74 * 74 will equal XXX(1)6

Therefore 72 is the answer. Check by multiplying it out.

Compare your answer with the correct one above

Question

If x and y are integers and xy + y2 is even, which of the following statements must be true?

I. 3y is odd

II. y/2 is an integer

III. xy is even

Answer

In order for the original statement to be true, the and $y^{2}$ terms must be either both odd or both even. Looking at each of the statements individually,

I. States that is odd, but only odd values multiplied by 3 are odd. If was an even number, the result would be even. But can be either odd or even, depending on what equals. Thus this statement COULD be true but does not HAVE to be true.

II. States that $\frac{y}{2}$ is an integer, and since only even numbers are cleanly divided by 2 (odd values result in a fraction) this ensures that is even. However, can also be odd, so this is a statement that COULD be true but does not HAVE to be true.

III. For exponents, only the base value determines whether it is even or odd - it does not indicate the value of y at all. Only even numbers raised to any power are even, thus, this ensures that is even. But can be odd as well, so this statement COULD be true but does not HAVE to be true.

An example of two integers that will work violate conditions II and III is and .

$(1)\cdot (3)+(3)^{2}=3+9=12$ , and even number.

$\frac{3}{2}$ is not an integer.

$1^{3}=1$ is not even.

Furthermore, any combination of 2 even integers will make the original statement true, and violate the Statement I.

Compare your answer with the correct one above

Question

Consider the inequality:

$x< x^{5}< x^{4}$

Which of the following could be a value of ?

Answer

Notice how $x^4$ is the greatest value. This often means that $x$ is negative as $(-1)^n=-1$ when $\dpi{100} n$ is odd and $(-1)^n=1$ when $\dpi{100} n$ is even.

Let us examine the first choice, $x=-\frac{3}{4}$

$x^5=-\frac{3^5}{4^5}=-\frac{243}{1024}> -\frac{3}{4}$

This can only be true of a negative value that lies between zero and one.

Compare your answer with the correct one above

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