Square Roots and Operations - PSAT Math

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

After doing so you are left with

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

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.

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

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

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

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.

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 values. Hence:

Do not try to simplify any further!

Begin by simplifying your more complex roots:

This lets us rewrite our expression:

Do the basic multiplications of coefficients:

Reorder the terms:

Finally, combine like terms:

Multiply the coefficents outside of the radicals.

Then multiply the radicans. Simplify by checking for a perfect square.

Final answer is your leading coefficent, , multiplied by the answer acquired by multiplying the terms under the radican, .

The final answer is .

Order of operations, first distributing the to all terms inside the parentheses.

The final answer is .

To square a number is to multiply that number by itself. Because 6 x 6 = 36 AND -6 x -6 = 36, both 6 and -6 are square roots of 36.

Multiplication of square roots is easy! You just have to multiply their contents by each other. Just don't forget to put the result "under" a square root! Therefore:

becomes

Now, you need to simplify this:

You can "pull out" two s. (Note, that it would be even easier to do this problem if you factor immediately instead of finding out that .)

After pulling out the s, you get:

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.

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

In order for the original statement to be true, the and 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 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 .

, and even number.

is not an integer.

is not even.

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

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

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

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

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

x=9

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

Simplify.

, so 243 is divisible by 3. .

, so 243 is not divisible by 7. - that is, .

, 243 is not a perfect square integer. - that is, .

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

Begin by squaring both sides of the equation:

Now solve for y:

Note that must be positive as defined in the original question. In this case, must be 12.