Simplifying Radicals - Algebra 2

Card 1 of 892

0

Didn't Know

0

Didn't Know

Knew It

0

1 of 2019 left

Question

$3\sqrt{48}-\sqrt{72}+5\sqrt{75}$

Tap to reveal answer

Answer

To add or subtract radicals, they must be the same root and have the same number under the radical before combining them. Look for perfect squares that divide into the number under the radicals because those can be simplified.

$3\sqrt{48}-\sqrt{72}+5\sqrt{75}$

$3\sqrt{16\cdot 3}-\sqrt{36\cdot 2}+5\sqrt{25\cdot 3}$

Take the square roots of each of the perfect squares in these radicals and bring it out of the radical. It will multiply to any coefficient in front of that radical

$3\cdot 4\sqrt{3}-6\sqrt{2}+5\cdot 5\sqrt{3}$

$12\sqrt{3}-6\sqrt{2}+25\sqrt{3}$

Remember, only radicals with the same number can be combined

$37\sqrt{3}-6\sqrt{2}$

This is the final answer.

← Didn't Know|Knew It →

Question

$What: is ; 3\sqrt{7} + 4\sqrt{7}, ?$

Tap to reveal answer

Answer

$Think : of; x = \sqrt7$

$Therefore, : 3\sqrt{7} + 4\sqrt{7} = 7\sqrt7$

← Didn't Know|Knew It →

Question

$\sqrt[3]{27} + \sqrt{64} =$

Tap to reveal answer

Answer

The third root of is

$\left ( 3\times 3\times 3 =27\right )$

and when added to the square root of 64, which is 8, you should get 11.

← Didn't Know|Knew It →

Question

Find the value of $\sqrt{80} + \sqrt{20}$ .

Tap to reveal answer

Answer

To solve this equation, we have to factor our radicals. We do this by finding numbers that multiply to give us the number within the radical.

$\sqrt{80} = \sqrt{20}\sqrt{4}$

$\sqrt{20} = \sqrt{5}\sqrt{4}$

Add them together:

$\sqrt5\sqrt4\sqrt4 + \sqrt5\sqrt4$

4 is a perfect square, so we can find the root:

$(2)(2)\sqrt5 + 2\sqrt5$

$4\sqrt5 + 2\sqrt5 = 6\sqrt5$

Since both have the same radical, we can combine them:

$4\sqrt5 + 2\sqrt5 = 6\sqrt5$

← Didn't Know|Knew It →

Question

Subtract and simplify:
$\small \frac{\sqrt3}{\sqrt5}-\frac{\sqrt3}{\sqrt2}$

Tap to reveal answer

Answer

$\small \frac{\sqrt3}{\sqrt5}-\frac{\sqrt3}{\sqrt2}$

Find the lease common denominator: $\small \sqrt{10}$

$\small \small \frac{\sqrt{3}}{\sqrt5}=\frac{\sqrt6}{\sqrt{10}}$

$\small \small \frac{\sqrt{3}}{\sqrt2}=\frac{\sqrt{15}}{\sqrt{10}}$

$\small \small \frac{\sqrt6}{\sqrt10}-\frac{\sqrt{15}}{\sqrt{10}}=\frac{\sqrt6-\sqrt{15}}{\sqrt{10}}$

A radical cannot be in the denominator:

$\small \small (\frac{\sqrt6-\sqrt{15}}{\sqrt{10}})(\frac{\sqrt{10}}{\sqrt{10}})=\frac{\sqrt{60}-\sqrt{150}}{10}$

$\small \sqrt{60}=2\sqrt{15}$

$\small \sqrt{150}=5\sqrt6$

$\small \frac{\sqrt{60}-\sqrt{150}}{10}=\frac{2\sqrt{15}-5\sqrt6}{10}$

← Didn't Know|Knew It →

Question

Simplify the following equation:

$f(x)=x^{2}-3\sqrt{4x}+4x^{\frac{1}{3}}-x-7x^{\frac{1}{2}}+5\sqrt[3]{x}$

Tap to reveal answer

Answer

When simplifying, you should always be on the lookout for like terms. While it might not look like there are like terms in , there are -- we just have to be able to rewrite it to see.

$f(x)=x^{2}-3\sqrt{4x}+4x^{\frac{1}{3}}-x-7x^{\frac{1}{2}}+5\sqrt[3]{x}$

$f(x)=x^{2}-3(4x)^{\frac{1}{2}}+4x^{\frac{1}{3}}-x-7x^{\frac{1}{2}}+5x^{\frac{1}{3}}$

Before we start combining terms, though, let's look a little more closely at this part:

$-3(4x)^{\frac{1}{2}}$

We need to "distribute" that exponent to everything in the parentheses, like so:

$-3\cdot 4^{\frac{1}{2}}\cdot x^{\frac{1}{2}}$

But 4 to the one-half power is just the square root of 4, or 2.

$-3\cdot 2\cdot x^{\frac{1}{2}}=-6x^{\frac{1}{2}}$

Okay, now let's see our equation.

$f(x)=x^{2}-6x^{\frac{1}{2}}+4x^{\frac{1}{3}}-x-7x^{\frac{1}{2}}+5x^{\frac{1}{3}}$

We need to start combining like terms. Take the terms that include x to the one-half power first.

$f(x)=x^{2}\mathbf{-6x^{\frac{1}{2}}}+4x^{\frac{1}{3}}-x\mathbf{-7x^{\frac{1}{2}}}+5x^{\frac{1}{3}}=x^{2}\mathbf{-13x^{\frac{1}{2}}}+4x^{\frac{1}{3}}-x+5x^{\frac{1}{3}}$

Now take the terms that have x to the one-third power.

$f(x)=x^{2}-13x^{\frac{1}{2}}\mathbf{+4x^{\frac{1}{3}}}-x\mathbf{+5x^{\frac{1}{3}}}=x^{2}-13x^{\frac{1}{2}}\mathbf{+9x^{\frac{1}{3}}}-x$

All that's left is to write them in order of descending exponents, then convert the fractional exponents into radicals (since that's what our answer choices look like).

$f(x)=x^{2}-x-13x^{\frac{1}{2}}+9x^{\frac{1}{3}}$

$f(x)=x^{2}-x-13\sqrt{x}+9\sqrt[3]{x}$

← Didn't Know|Knew It →

Question

Simplify.

$2\sqrt{16} - 4\sqrt{8}$

Tap to reveal answer

Answer

We can simplify both radicals:

$\sqrt{16} =4$ and $\sqrt{8}=\sqrt{4\times 2} = 2\sqrt{2}$

Plug in the simplified radicals into the equation:

$2(4) - 4(2\sqrt{2})$

which leaves us with:

$8- 8\sqrt{2}$

Because these are not like terms, we cannot simplify this further.

← Didn't Know|Knew It →

Question

Simplify.

$-8\sqrt{27}-11\sqrt{9}+2\sqrt{3}$

Tap to reveal answer

Answer

Only the first two radicals can be simplified:

$\sqrt{27}=\sqrt{9\times 3}=3\sqrt{3}$ and $\sqrt{9}=3$

Plug in the simplified radicals into the equation:

$-8(3\sqrt{3})-11(3)+2\sqrt{3}$

$-24\sqrt{3}-33+2\sqrt{3}$

We can now simplify the equation by combining the like terms:

$-22\sqrt{3}-33$

← Didn't Know|Knew It →

Question

Simplify.

$5\sqrt{28} +\sqrt{7} -7\sqrt{63}$

Tap to reveal answer

Answer

We can simplify the first and third radicals:

$\sqrt{28} =\sqrt{7\times 4} =2\sqrt{7}$ and $\sqrt{63} =\sqrt{9\times 7 } =3\sqrt{7}$

Plug in the simplified radicals into the equation:

$5(2\sqrt{7}) + \sqrt{7} -7(3\sqrt{7})$

$10\sqrt{7} + \sqrt{7 } -21\sqrt{7}$

Combine the like terms:

$-10\sqrt{7}$

← Didn't Know|Knew It →

Question

Solve.

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

Tap to reveal answer

Answer

When adding and subtracting radicals, make the sure radicand or inside the square root are the same.

If they are the same, just add the numbers in front of the radical.

Since they are not the same, the answer is just the problem stated.

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

← Didn't Know|Knew It →

Question

Solve.

$\sqrt{7}-\sqrt{4}$

Tap to reveal answer

Answer

When adding and subtracting radicals, make the sure radicand or inside the square root are the same.

If they are the same, just add the numbers in front of the radical.

Since they are not the same, the answer is just the problem stated.

$\sqrt{7}-\sqrt{4}$

← Didn't Know|Knew It →

Question

Solve.

$\sqrt{9}+\sqrt{4}+\sqrt{1}$

Tap to reveal answer

Answer

When adding and subtracting radicals, make the sure radicand or inside the square root are the same.

If they are the same, just add the numbers in front of the radical.

Looking carefully at the radicand we will notice that each radicand is a perfect sqare. This means we are able to reduce the radical into an integer.

and are all sqaure numbers so instead we have a simple algebraic problem:

$\sqrt9+\sqrt4 +\sqrt1=\sqrt{3\cdot 3}+\sqrt{2\cdot 2}+\sqrt{1\cdot 1}=3+2+1$ which the answer is .

← Didn't Know|Knew It →

Question

Solve.

$\sqrt{49}-\sqrt{16}-\sqrt{9}$

Tap to reveal answer

Answer

When adding and subtracting radicals, make the sure radicand or inside the square root are the same.

If they are the same, just add the numbers in front of the radical.

Looking carefully at the radicand we will see that each radicand is a perfect square. Therefore, we are able to reduce all of the radicals into simple integers.

and are all sqaure numbers so instead we have a simple algebraic problem:

$\sqrt{49}-\sqrt{16}-\sqrt9=\sqrt{7\cdot 7}-\sqrt{4\cdot 4}-\sqrt{3\cdot 3}=7-4-3$ which the answer is .

← Didn't Know|Knew It →

Question

Solve.

$\sqrt{3}+5\sqrt{3}+8\sqrt{3}$

Tap to reveal answer

Answer

When adding and subtracting radicals, make the sure radicand or inside the square root are the same.

If they are the same, just add the numbers in front of the radical. If they are not the same, the answer is just the problem stated.

Since they are the same, just add the numbers in front of the radical: which is

Therefore, our final answer is the sum of the integers and the radical:

$14\sqrt3$

← Didn't Know|Knew It →

Question

Solve.

$20\sqrt{5}-13\sqrt{5}-\sqrt{5}$

Tap to reveal answer

Answer

When adding and subtracting radicals, make the sure radicand or inside the square root are the same.

If they are the same, just add the numbers in front of the radical.

If they are not the same, the answer is just the problem stated.

Since they are the same, just subtract the numbers in front: which is

Therefore, our final answer is this sum with the radical added to the end:

$6\sqrt5$

← Didn't Know|Knew It →

Question

Solve.

$\sqrt{5}+6\sqrt{5}-4\sqrt{5}$

Tap to reveal answer

Answer

When adding and subtracting radicals, make the sure radicand or inside the square root are the same.

If they are the same, just add the numbers in front of the radical.

If they are not the same, the answer is just the problem stated.

Since they are the same, just add and subtract the numbers in front: which is

Therefore, the final answer will be this sum and the radical added to the end:

$3\sqrt5$

← Didn't Know|Knew It →

Question

Solve.

$\sqrt{8}+\sqrt{2}$

Tap to reveal answer

Answer

When adding and subtracting radicals, make the sure radicand or inside the square root are the same.

If they are the same, just add the numbers in front of the radical.

If they are not the same, the answer is just the problem stated.

Even though it's not the same, double check you can simplify the radicand. Look for perfect squares. Since $4\cdot 2 = 8$ and is a perfect square, we can rewrite like this: $\sqrt{8}=\sqrt{4\cdot 2}=2\sqrt{2}$ .

Now we have the same radicand, we can now add them easily to get $3\sqrt{2}$ .

← Didn't Know|Knew It →

Question

Solve.

$\sqrt{75}-2\sqrt{3}$

Tap to reveal answer

Answer

When adding and subtracting radicals, make the sure radicand or inside the square root are the same. If they are the same, just add the numbers in front of the radical. If they are not the same, the answer is just the problem stated.

Even though it's not the same, double check you can simplify the radicand. Look for perfect squares.

Since $25\cdot 3 = 75$ and is a perfect square, we can rewrite like this: $\sqrt{75}=\sqrt{25\cdot 3}=5\sqrt{3}$ .

Now we have the same radicand, we can now subtract them easily to get $3\sqrt{3}$ .

← Didn't Know|Knew It →

Question

Solve.

$\sqrt{200}+\sqrt{50}$

Tap to reveal answer

Answer

When adding and subtracting radicals, make the sure radicand or inside the square root are the same. If they are the same, just add the numbers in front of the radical. If they are not the same, the answer is just the problem stated.

Even though it's not the same, double check you can simplify the radicand. Look for perfect squares.

Since $100\cdot 2 = 200$ , $25\cdot 2=50$ and and are perfect squares, we can rewrite like this:

$\sqrt{200}=\sqrt{100\cdot 2}=10\sqrt{2}$ and $\sqrt{50}=\sqrt{25\cdot 2}=5\sqrt{2}$ .

Now that we have the same radicand, we can add them easily to get:

$10\sqrt{2}+5\sqrt{2}$

$15\sqrt{2}$ .

← Didn't Know|Knew It →

Question

Solve.

$2\sqrt{100}+8\sqrt{16}-5\sqrt{175}$

Tap to reveal answer

Answer

When adding and subtracting radicals, make the sure radicand or inside the square root are the same. If they are the same, just add the numbers in front of the radical. If they are not the same, the answer is just the problem stated.

Even though it's not the same, double check you can simplify the radicand. Look for perfect squares. and are perfect squares so we can rewrite like this:

$2\sqrt{100}=2\sqrt{10\cdot 10}=2\cdot 10=20$

$8\sqrt{16}=8\sqrt{4\cdot 4}=8\cdot4=32$

Since $25\cdot 7 = 175$ and is a perfect square, we can rewrite like this:

$5\sqrt{175}=5\sqrt{25\cdot7}=5\cdot5\sqrt{7}=25\sqrt{7}$ .

Lets add everything up.

$20+32-25\sqrt{7}$ $=52-25\sqrt{7}$ .

The reason this is the answer, because the is associated with the radical and we can't subtract a whole number with the radical. They are not the same.

← Didn't Know|Knew It →

0

Knew It