Square Roots and Operations - PSAT Math

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Question

If $\sqrt{x}=3^2$ what is $x$ ?

Answer

Square both sides:

x = (32)2 = 92 = 81

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Question

Simplify.

$\sqrt{32}+\sqrt{64}+\sqrt{8}$

Answer

$\sqrt{32}+\sqrt{64}+\sqrt{8}$

First step is to find perfect squares in all of our radicans.

$\sqrt{32}\rightarrow\sqrt{16\cdot 2}\rightarrow4\sqrt{2}$

$\sqrt{64}\rightarrow\sqrt{8\cdot 8}\rightarrow8$

$\sqrt{8}\rightarrow\sqrt{4\cdot 2}\rightarrow2\sqrt{2}$

After doing so you are left with $4\sqrt{2}+2\sqrt{2}+8$

Just like fractions you can only add together coefficents with like terms under the radical.

$6\sqrt{2}+8$

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Question

Simplify:

$\sqrt{27}-\sqrt{48}+\sqrt{81}$

Answer

To combine radicals, they must have the same radicand. Therefore, we must find the perfect squares in each of our square roots and pull them out.

$\sqrt{27}\rightarrow \sqrt{93}\rightarrow \sqrt{3^23}\rightarrow 3\sqrt{3}$

$\sqrt{48}\rightarrow \sqrt{163}\rightarrow \sqrt{4^23}\rightarrow 4\sqrt{3}$

$\sqrt{81}\rightarrow\sqrt{9*9}\rightarrow\sqrt{9^2}\rightarrow9$

Now, we plug these equivalent expressions back into our equation and simplify:

$\sqrt{27}-\sqrt{48}+\sqrt{81}$

$3\sqrt3-4\sqrt3+9$

$-\sqrt3+9$

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Question

Simplify:

$\sqrt{54} - \sqrt{24} + \sqrt{6}$

Answer

Simplify each of the radicals by factoring out a perfect square:

$\sqrt{54} - \sqrt{24} + \sqrt{6}$

$= \sqrt{9 \cdot 6} - \sqrt{4\cdot 6} + \sqrt{6}$

$= \sqrt{9} \cdot \sqrt{6} - \sqrt{4} \cdot \sqrt{6} + \sqrt{6}$

$= 3\sqrt{6} - 2 \sqrt{6} + 1\sqrt{6}$

$=\left ( 3 - 2 + 1 \right ) \sqrt{6}$

$= 2 \sqrt{6}$

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Question

Simplify the expression:

$\sqrt{45} + \sqrt{125} + \sqrt{175}$

Answer

For each of the expressions, factor out a perfect square:

$\sqrt{45} + \sqrt{125} + \sqrt{175}$

$= \sqrt{9 \cdot 5} + \sqrt{25 \cdot 5} + \sqrt{25 \cdot 7}$

$= \sqrt{9 } \cdot \sqrt{ 5} + \sqrt{25 } \cdot \sqrt{ 5}+ \sqrt{25} \cdot \sqrt{ 7}$

$=3 \sqrt{ 5} +5 \sqrt{ 5}+5 \sqrt{ 7}$

$=\left (3 +5 \right ) \sqrt{ 5}+5 \sqrt{ 7}$

$=8 \sqrt{ 5}+5 \sqrt{ 7}$

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Question

Add the square roots into one term:

$\small {\sqrt{12}} + {\sqrt{75}}$

Answer

In order to solve this problem we need to simplfy each of the radicals. By doing this we will get two terms that have the same number under the radical which will allow us to combine the terms.

$\small {\sqrt{12}} + {\sqrt{75}}$

$\small = \small {\sqrt{4}} \times {\sqrt{3}} + {\sqrt{25}} \times {\sqrt{3}}$

$\small \small = 2 {\sqrt{3}} + 5 {\sqrt{3}}$

$\small = 7 {\sqrt{3}}$

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Question

Simplify:

$\sqrt{2}+\sqrt{2}+\sqrt{3}+3\sqrt{5}$

Answer

Remember that you treat square roots like you do variables in the sense that you just add the like factors. In this problem, the only set of like factors is the pair of $\sqrt{2}$ values. Hence:

$\sqrt{2}+\sqrt{2}+\sqrt{3}+3\sqrt{5} = 2\sqrt{2}+\sqrt{3}+3\sqrt{5}$

Do not try to simplify any further!

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Question

Simplify:

$\sqrt{2}+3\sqrt{12}+2\sqrt{50}+\sqrt{3}$

Answer

Begin by simplifying your more complex roots:

$\sqrt{12}= \sqrt{223}=2\sqrt{3}$

$\sqrt{50} = \sqrt{552}=5\sqrt{2}$

This lets us rewrite our expression:

$\sqrt{2}+3\sqrt{12}+2\sqrt{50}+\sqrt{3} = \sqrt{2}+3(2\sqrt{3})+2(5\sqrt{2})+\sqrt{3}$

Do the basic multiplications of coefficients:

$\sqrt{2}+3(2\sqrt{3})+2(5\sqrt{2})+\sqrt{3} = \sqrt{2}+6\sqrt{3}+10\sqrt{2}+\sqrt{3}$

Reorder the terms:

$\sqrt{2}+10\sqrt{2}+6\sqrt{3}+\sqrt{3}$

Finally, combine like terms:

$11\sqrt{2}+7\sqrt{3}$

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Question

(√27 + √12) / √3 is equal to

Answer

√27 is the same as 3√3, while √12 is the same as 2√3.

3√3 + 2√3 = 5√3

(5√3)/(√3) = 5

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Question

Divide and simplify. Assume all integers are positive real numbers.

$\sqrt{\frac{32}{4}}$

Answer

$\sqrt{\frac{32}{4}}$

There are two ways to solve this problem. First you can divide the numbers under the radical. Then simplify.

Example 1

$\sqrt{\frac{32}{4}}=\sqrt{8}=\sqrt{4}\cdot\sqrt{2}=2\sqrt{2}$

Example 2

Find the square root of both numerator and denominator, simplifying as much as possible then dividing out like terms.

$\sqrt{\frac{32}{4}}=\frac{\sqrt{32}}{\sqrt{4}}=\frac{\sqrt{16}\cdot \sqrt{2}}{\sqrt{4}}=\frac{4\sqrt{2}}{2}=2\sqrt{2}$

Both methods will give you the correct answer of $2\sqrt{2}$ .

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Question

Simplify:

$\sqrt{\frac{64}{169}}$

Answer

To simplfy, we must first distribute the square root.

$\frac{\sqrt{64}}{\sqrt{169}}$

Next, we can simplify each of the square roots.

$\frac{8}{13}$

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Question

Find the quotient:

$\frac{\sqrt{54}}{\sqrt{45}}$

Answer

Simplify each radical:

$\frac{\sqrt{54}}{\sqrt{45}} = \frac{\sqrt{9\cdot 6}}{\sqrt{9\cdot 5}}=\frac{3\sqrt{6}}{3\sqrt{5}}=\frac{\sqrt{6}}{\sqrt{5}}$

Rationalize the denominator:

$\frac{\sqrt{6}}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}}= \frac{\sqrt{30}}{5}$

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Question

Find the quotient:

$\sqrt{\frac{18}{2}}$

Answer

Find the quotient:

$\frac{\sqrt{18}}{\sqrt{2}}$

There are two ways to approach this problem.

Option 1: Combine the radicals first, the reduce

$\frac{\sqrt{18}}{\sqrt{2}}=\sqrt{\frac{18}{2}}=\sqrt{9}=3$

Option 2: Simplify the radicals first, then reduce

$\frac{\sqrt{18}}{\sqrt{2}}=\frac{\sqrt{9\cdot 2}}{\sqrt{2}}=\frac{3\sqrt{2}}{\sqrt{2}}=3$

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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.

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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.

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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.

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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

Simplify.

$\sqrt{\frac{16}{121}}$

Answer

$\sqrt{\frac{16}{121}}$

Take the square root of both the top and bottom terms.

$\frac{\sqrt{16}}{\sqrt{121}}$

Simplify.

$\frac{4}{11}$

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Question

Let the universal set be the set of all positive integers.

Let be the set of all multiples of 3; let be the set of all multiples of 7; let be the set of all perfect square integers. Which of the following statements is true of 243?

Note: means "the complement of ", etc.

Answer

$243 \div 3 = 81$ , so 243 is divisible by 3. $243 \in A$ .

$243 \div 7= 34 \textup{ R }5$ , so 243 is not divisible by 7. - that is, $243 \in B'$ .

$15 = \sqrt{225}< \sqrt{243} < \sqrt{256} = 16$ , 243 is not a perfect square integer. - that is, $243 \in C'$ .

Since 243 is an element of , , and , it is an element of their intersection. The correct choice is that

$243 \in A \cap B' \cap C'$

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