QUESTION IMAGE
Question
- \\(sqrt{8x^7y}\\) 10) \\(sqrt{54x^3y^8}\\) 11) \\(sqrt{90x^{10}y^{20}}\\) 12) \\(sqrt{172x^5y^{13}}\\) 13) \\(sqrt3{8x^7y}\\) 14) \\(sqrt3{54x^3y^8}\\) 15) \\(sqrt3{90x^{10}y^{20}}\\) 16) \\(sqrt3{135x^5y^{13}}\\)
Let's solve these radical simplification problems one by one. We'll start with the square roots (problems 9 - 12) and then the cube roots (problems 13 - 16).
Problem 9: Simplify $\boldsymbol{\sqrt{8x^7y}}$
Step 1: Factor the radicand into perfect squares and other factors
We know that $8 = 4 \times 2$ (where $4$ is a perfect square), $x^7 = x^6 \times x$ (where $x^6$ is a perfect square since $6$ is divisible by $2$), and $y$ remains as is (assuming $y$ has no perfect square factors for now).
So, $\sqrt{8x^7y} = \sqrt{4 \times 2 \times x^6 \times x \times y}$
Step 2: Use the property $\sqrt{ab} = \sqrt{a} \times \sqrt{b}$
$\sqrt{4 \times 2 \times x^6 \times x \times y} = \sqrt{4} \times \sqrt{x^6} \times \sqrt{2xy}$
Step 3: Simplify the perfect square roots
$\sqrt{4} = 2$ and $\sqrt{x^6} = x^3$ (since $\sqrt{x^n} = x^{\frac{n}{2}}$, so $x^{\frac{6}{2}} = x^3$)
So, putting it together, we get $2x^3\sqrt{2xy}$
Problem 10: Simplify $\boldsymbol{\sqrt{54x^3y^8}}$
Step 1: Factor the radicand into perfect squares and other factors
$54 = 9 \times 6$ (where $9$ is a perfect square), $x^3 = x^2 \times x$ (where $x^2$ is a perfect square), and $y^8$ is a perfect square (since $8$ is divisible by $2$).
So, $\sqrt{54x^3y^8} = \sqrt{9 \times 6 \times x^2 \times x \times y^8}$
Step 2: Use the property $\sqrt{ab} = \sqrt{a} \times \sqrt{b}$
$\sqrt{9 \times 6 \times x^2 \times x \times y^8} = \sqrt{9} \times \sqrt{x^2} \times \sqrt{y^8} \times \sqrt{6x}$
Step 3: Simplify the perfect square roots
$\sqrt{9} = 3$, $\sqrt{x^2} = x$, and $\sqrt{y^8} = y^4$ (since $\sqrt{y^n} = y^{\frac{n}{2}}$, so $y^{\frac{8}{2}} = y^4$)
Putting it together, we get $3xy^4\sqrt{6x}$
Problem 11: Simplify $\boldsymbol{\sqrt{90x^{10}y^{20}}}$
Step 1: Factor the radicand into perfect squares and other factors
$90 = 9 \times 10$ (where $9$ is a perfect square), $x^{10}$ is a perfect square (since $10$ is divisible by $2$), and $y^{20}$ is a perfect square (since $20$ is divisible by $2$).
So, $\sqrt{90x^{10}y^{20}} = \sqrt{9 \times 10 \times x^{10} \times y^{20}}$
Step 2: Use the property $\sqrt{ab} = \sqrt{a} \times \sqrt{b}$
$\sqrt{9 \times 10 \times x^{10} \times y^{20}} = \sqrt{9} \times \sqrt{x^{10}} \times \sqrt{y^{20}} \times \sqrt{10}$
Step 3: Simplify the perfect square roots
$\sqrt{9} = 3$, $\sqrt{x^{10}} = x^5$ (since $\sqrt{x^n} = x^{\frac{n}{2}}$, so $x^{\frac{10}{2}} = x^5$), and $\sqrt{y^{20}} = y^{10}$ (since $\sqrt{y^n} = y^{\frac{n}{2}}$, so $y^{\frac{20}{2}} = y^{10}$)
Putting it together, we get $3x^5y^{10}\sqrt{10}$
Problem 12: Simplify $\boldsymbol{\sqrt{172x^5y^{13}}}$
Step 1: Factor the radicand into perfect squares and other factors
$172 = 4 \times 43$ (where $4$ is a perfect square), $x^5 = x^4 \times x$ (where $x^4$ is a perfect square), and $y^{13} = y^{12} \times y$ (where $y^{12}$ is a perfect square).
So, $\sqrt{172x^5y^{13}} = \sqrt{4 \times 43 \times x^4 \times x \times y^{12} \times y}$
Step 2: Use the property $\sqrt{ab} = \sqrt{a} \times \sqrt{b}$
$\sqrt{4 \times 43 \times x^4 \times x \times y^{12} \times y} = \sqrt{4} \times \sqrt{x^4} \times \sqrt{y^{12}} \times \sqrt{43xy}$
Step 3: Simplify the perfect square roots
$\sqrt{4} = 2$, $\sqrt{x^4} = x^2$, and $\sqrt{y^{12}} = y^6$ (since $\sqrt{y^n} = y^{\frac{n}{2}}$, so $y^{\frac{12}{2}} = y^6$)
Putting it together, we get $2x^2y^6\sqrt{43xy}$
Problem 13: Simplify $\boldsymbol{\sqrt[3]{8x^7y}}$
Step 1: Factor the radicand into perfect cubes and other factors
We know that $8 = 2^3$ (a perfect c…
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Let's solve these radical simplification problems one by one. We'll start with the square roots (problems 9 - 12) and then the cube roots (problems 13 - 16).
Problem 9: Simplify $\boldsymbol{\sqrt{8x^7y}}$
Step 1: Factor the radicand into perfect squares and other factors
We know that $8 = 4 \times 2$ (where $4$ is a perfect square), $x^7 = x^6 \times x$ (where $x^6$ is a perfect square since $6$ is divisible by $2$), and $y$ remains as is (assuming $y$ has no perfect square factors for now).
So, $\sqrt{8x^7y} = \sqrt{4 \times 2 \times x^6 \times x \times y}$
Step 2: Use the property $\sqrt{ab} = \sqrt{a} \times \sqrt{b}$
$\sqrt{4 \times 2 \times x^6 \times x \times y} = \sqrt{4} \times \sqrt{x^6} \times \sqrt{2xy}$
Step 3: Simplify the perfect square roots
$\sqrt{4} = 2$ and $\sqrt{x^6} = x^3$ (since $\sqrt{x^n} = x^{\frac{n}{2}}$, so $x^{\frac{6}{2}} = x^3$)
So, putting it together, we get $2x^3\sqrt{2xy}$
Problem 10: Simplify $\boldsymbol{\sqrt{54x^3y^8}}$
Step 1: Factor the radicand into perfect squares and other factors
$54 = 9 \times 6$ (where $9$ is a perfect square), $x^3 = x^2 \times x$ (where $x^2$ is a perfect square), and $y^8$ is a perfect square (since $8$ is divisible by $2$).
So, $\sqrt{54x^3y^8} = \sqrt{9 \times 6 \times x^2 \times x \times y^8}$
Step 2: Use the property $\sqrt{ab} = \sqrt{a} \times \sqrt{b}$
$\sqrt{9 \times 6 \times x^2 \times x \times y^8} = \sqrt{9} \times \sqrt{x^2} \times \sqrt{y^8} \times \sqrt{6x}$
Step 3: Simplify the perfect square roots
$\sqrt{9} = 3$, $\sqrt{x^2} = x$, and $\sqrt{y^8} = y^4$ (since $\sqrt{y^n} = y^{\frac{n}{2}}$, so $y^{\frac{8}{2}} = y^4$)
Putting it together, we get $3xy^4\sqrt{6x}$
Problem 11: Simplify $\boldsymbol{\sqrt{90x^{10}y^{20}}}$
Step 1: Factor the radicand into perfect squares and other factors
$90 = 9 \times 10$ (where $9$ is a perfect square), $x^{10}$ is a perfect square (since $10$ is divisible by $2$), and $y^{20}$ is a perfect square (since $20$ is divisible by $2$).
So, $\sqrt{90x^{10}y^{20}} = \sqrt{9 \times 10 \times x^{10} \times y^{20}}$
Step 2: Use the property $\sqrt{ab} = \sqrt{a} \times \sqrt{b}$
$\sqrt{9 \times 10 \times x^{10} \times y^{20}} = \sqrt{9} \times \sqrt{x^{10}} \times \sqrt{y^{20}} \times \sqrt{10}$
Step 3: Simplify the perfect square roots
$\sqrt{9} = 3$, $\sqrt{x^{10}} = x^5$ (since $\sqrt{x^n} = x^{\frac{n}{2}}$, so $x^{\frac{10}{2}} = x^5$), and $\sqrt{y^{20}} = y^{10}$ (since $\sqrt{y^n} = y^{\frac{n}{2}}$, so $y^{\frac{20}{2}} = y^{10}$)
Putting it together, we get $3x^5y^{10}\sqrt{10}$
Problem 12: Simplify $\boldsymbol{\sqrt{172x^5y^{13}}}$
Step 1: Factor the radicand into perfect squares and other factors
$172 = 4 \times 43$ (where $4$ is a perfect square), $x^5 = x^4 \times x$ (where $x^4$ is a perfect square), and $y^{13} = y^{12} \times y$ (where $y^{12}$ is a perfect square).
So, $\sqrt{172x^5y^{13}} = \sqrt{4 \times 43 \times x^4 \times x \times y^{12} \times y}$
Step 2: Use the property $\sqrt{ab} = \sqrt{a} \times \sqrt{b}$
$\sqrt{4 \times 43 \times x^4 \times x \times y^{12} \times y} = \sqrt{4} \times \sqrt{x^4} \times \sqrt{y^{12}} \times \sqrt{43xy}$
Step 3: Simplify the perfect square roots
$\sqrt{4} = 2$, $\sqrt{x^4} = x^2$, and $\sqrt{y^{12}} = y^6$ (since $\sqrt{y^n} = y^{\frac{n}{2}}$, so $y^{\frac{12}{2}} = y^6$)
Putting it together, we get $2x^2y^6\sqrt{43xy}$
Problem 13: Simplify $\boldsymbol{\sqrt[3]{8x^7y}}$
Step 1: Factor the radicand into perfect cubes and other factors
We know that $8 = 2^3$ (a perfect cube), $x^7 = x^6 \times x$ (where $x^6$ is a perfect cube since $6$ is divisible by $3$), and $y$ remains as is (assuming $y$ has no perfect cube factors for now).
So, $\sqrt[3]{8x^7y} = \sqrt[3]{2^3 \times x^6 \times x \times y}$
Step 2: Use the property $\sqrt[3]{ab} = \sqrt[3]{a} \times \sqrt[3]{b}$
$\sqrt[3]{2^3 \times x^6 \times x \times y} = \sqrt[3]{2^3} \times \sqrt[3]{x^6} \times \sqrt[3]{xy}$
Step 3: Simplify the perfect cube roots
$\sqrt[3]{2^3} = 2$ and $\sqrt[3]{x^6} = x^2$ (since $\sqrt[3]{x^n} = x^{\frac{n}{3}}$, so $x^{\frac{6}{3}} = x^2$)
Putting it together, we get $2x^2\sqrt[3]{xy}$
Problem 14: Simplify $\boldsymbol{\sqrt[3]{54x^3y^8}}$
Step 1: Factor the radicand into perfect cubes and other factors
$54 = 27 \times 2$ (where $27$ is a perfect cube, $27 = 3^3$), $x^3$ is a perfect cube, and $y^8 = y^6 \times y^2$ (where $y^6$ is a perfect cube since $6$ is divisible by $3$).
So, $\sqrt[3]{54x^3y^8} = \sqrt[3]{27 \times 2 \times x^3 \times y^6 \times y^2}$
Step 2: Use the property $\sqrt[3]{ab} = \sqrt[3]{a} \times \sqrt[3]{b}$
$\sqrt[3]{27 \times 2 \times x^3 \times y^6 \times y^2} = \sqrt[3]{27} \times \sqrt[3]{x^3} \times \sqrt[3]{y^6} \times \sqrt[3]{2y^2}$
Step 3: Simplify the perfect cube roots
$\sqrt[3]{27} = 3$, $\sqrt[3]{x^3} = x$, and $\sqrt[3]{y^6} = y^2$ (since $\sqrt[3]{y^n} = y^{\frac{n}{3}}$, so $y^{\frac{6}{3}} = y^2$)
Putting it together, we get $3xy^2\sqrt[3]{2y^2}$
Problem 15: Simplify $\boldsymbol{\sqrt[3]{90x^{10}y^{20}}}$
Step 1: Factor the radicand into perfect cubes and other factors
$90 = 27 \times \frac{10}{3}$? Wait, no. Wait, $90 = 27 \times \frac{10}{3}$ is not helpful. Wait, $90 = 27 \times \frac{10}{3}$ is wrong. Wait, $90 = 27 \times 3 + 9$, no. Wait, let's factor $90$ as $27 \times \frac{10}{3}$ no, that's not integer. Wait, actually, $90 = 27 \times \frac{10}{3}$ is incorrect. Wait, $90 = 27 \times 3 + 9$, no. Wait, maybe I made a mistake. Wait, $90 = 27 \times 3 + 9$ is not helpful. Wait, $90 = 27 \times \frac{10}{3}$ is wrong. Wait, let's factor $90$ into prime factors: $90 = 2 \times 3^3 \times 5$. Ah! There we go. So $90 = 3^3 \times 10$ (since $3^3 = 27$ and $27 \times 10 = 270$? Wait no, $3^3 = 27$, $27 \times 3 = 81$, $81 + 9 = 90$. Wait, no, $90 = 2 \times 3^2 \times 5$. Oh! I see, I messed up. $3^3 = 27$, $27 \times 3 = 81$, $81 + 9 = 90$, but $9 = 3^2$. So $90 = 2 \times 3^3 \times 5$? Wait, $3^3 = 27$, $27 \times 2 = 54$, $54 + 36 = 90$, no. Wait, prime factorization of $90$: $90 = 2 \times 3^2 \times 5$. So there is no perfect cube factor in $90$ except $1$? Wait, no, $3^3 = 27$, $27$ is a factor of $90$? $90 \div 27 = 3.333...$ No, $27 \times 3 = 81$, $27 \times 4 = 108$. So $90$ has no perfect cube factors except $1$. Wait, but $x^{10} = x^9 \times x$ (where $x^9$ is a perfect cube since $9$ is divisible by $3$), and $y^{20} = y^{18} \times y^2$ (where $y^{18}$ is a perfect cube since $18$ is divisible by $3$).
So, $\sqrt[3]{90x^{10}y^{20}} = \sqrt[3]{90 \times x^9 \times x \times y^{18} \times y^2}$
Step 2: Use the property $\sqrt[3]{ab} = \sqrt[3]{a} \times \sqrt[3]{b}$
$\sqrt[3]{90 \times x^9 \times x \times y^{18} \times y^2} = \sqrt[3]{x^9} \times \sqrt[3]{y^{18}} \times \sqrt[3]{90xy^2}$
Step 3: Simplify the perfect cube roots
$\sqrt[3]{x^9} = x^3$ (since $\sqrt[3]{x^n} = x^{\frac{n}{3}}$, so $x^{\frac{9}{3}} = x^3$) and $\sqrt[3]{y^{18}} = y^6$ (since $\sqrt[3]{y^n} = y^{\frac{n}{3}}$, so $y^{\frac{18}{3}} = y^6$)
Putting it together, we get $x^3y^6\sqrt[3]{90xy^2}$
Problem 16: Simplify $\boldsymbol{\sqrt[3]{135x^5y^{13}}}$
Step 1: Factor the radicand into perfect cubes and other factors
$135 = 27 \times 5$ (where $27$ is a perfect cube, $27 = 3^3$), $x^5 = x^3 \times x^2$ (where $x^3$ is a perfect cube), and $y^{13} = y^{12} \times y$ (where $y^{12}$ is a perfect cube since $12$ is divisible by $3$).
So, $\sqrt[3]{135x^5y^{13}} = \sqrt[3]{27 \times 5 \times x^3 \times x^2 \times y^{12} \times y}$
Step 2: Use the property $\sqrt[3]{ab} = \sqrt[3]{a} \times \sqrt[3]{b}$
$\sqrt[3]{27 \times 5 \times x^3 \times x^2 \times y^{12} \times y} = \sqrt[3]{27} \times \sqrt[3]{x^3} \times \sqrt[3]{y^{12}} \times \sqrt[3]{5x^2y}$
Step 3: Simplify the perfect cube roots
$\sqrt[3]{27} = 3$, $\sqrt[3]{x^3} = x$, and $\sqrt[3]{y^{12}} = y^4$ (since $\sqrt[3]{y^n} = y^{\frac{n}{3}}$, so $y^{\frac{12}{3}} = y^4$)
Putting it together, we get $3xy^4\sqrt[3]{5x^2y}$
Final Answers:
- $\boldsymbol{2x^3\sqrt{2xy}}$
- $\boldsymbol{3xy^4\sqrt{6x}}$
- $\boldsymbol{3x^5y^{10}\sqrt{10}}$
- $\boldsymbol{2x^2y^6\sqrt{43xy}}$
- $\boldsymbol{2x^2\sqrt[3]{xy}}$
- $\boldsymbol{3xy^2\sqrt[3]{2y^2}}$
- $\boldsymbol{x^3y^6\sqrt[3]{90xy^2}}$ (Wait, no, earlier we had $90$ with no perfect cube factors, so actually, $\sqrt[3]{90x^{10}y^{20}} = \sqrt[3]{x^9 \times y^{18} \times 90xy^2} = x^3y^6\sqrt[3]{90xy^2}$)
- $\boldsymbol{3xy^4\sqrt[3]{5x^2y}}$