Asked by mr alpha
Complete the expression subtracting the radicals
14 sqrt3 - 4 sqrt⬜ = 2 sqrt⬜
14 sqrt3 - 4 sqrt⬜ = 2 sqrt⬜
Answers
Answered by
mr alpha
What is the product of the following in simplest radical form?
√85× 3√32
√85× 3√32
Answered by
GPT-4o mini
To complete the expression \( 14\sqrt{3} - 4\sqrt{\square} = 2\sqrt{\square} \), we first want to isolate the radical term.
Starting from the equation:
\[
14\sqrt{3} - 4\sqrt{\square} = 2\sqrt{\square}
\]
we can add \( 4\sqrt{\square} \) to both sides:
\[
14\sqrt{3} = 2\sqrt{\square} + 4\sqrt{\square}
\]
This simplifies to:
\[
14\sqrt{3} = 6\sqrt{\square}
\]
Next, we isolate \( \sqrt{\square} \) by dividing both sides by \( 6 \):
\[
\sqrt{\square} = \frac{14\sqrt{3}}{6}
\]
Simplifying the right side:
\[
\sqrt{\square} = \frac{7\sqrt{3}}{3}
\]
Now, to get \( \square \), we square both sides:
\[
\square = \left(\frac{7\sqrt{3}}{3}\right)^2
\]
Calculating the square:
\[
\square = \frac{49 \cdot 3}{9} = \frac{147}{9} = \frac{49}{3}
\]
Thus, completing the expression yields:
\[
\square = \frac{49}{3}
\]
So the full expression is:
\[
14\sqrt{3} - 4\sqrt{\frac{49}{3}} = 2\sqrt{\frac{49}{3}}
\]
Starting from the equation:
\[
14\sqrt{3} - 4\sqrt{\square} = 2\sqrt{\square}
\]
we can add \( 4\sqrt{\square} \) to both sides:
\[
14\sqrt{3} = 2\sqrt{\square} + 4\sqrt{\square}
\]
This simplifies to:
\[
14\sqrt{3} = 6\sqrt{\square}
\]
Next, we isolate \( \sqrt{\square} \) by dividing both sides by \( 6 \):
\[
\sqrt{\square} = \frac{14\sqrt{3}}{6}
\]
Simplifying the right side:
\[
\sqrt{\square} = \frac{7\sqrt{3}}{3}
\]
Now, to get \( \square \), we square both sides:
\[
\square = \left(\frac{7\sqrt{3}}{3}\right)^2
\]
Calculating the square:
\[
\square = \frac{49 \cdot 3}{9} = \frac{147}{9} = \frac{49}{3}
\]
Thus, completing the expression yields:
\[
\square = \frac{49}{3}
\]
So the full expression is:
\[
14\sqrt{3} - 4\sqrt{\frac{49}{3}} = 2\sqrt{\frac{49}{3}}
\]
Answered by
GPT-4o mini
To find the product of \( \sqrt{85} \times 3\sqrt{32} \), we can first separate the constants from the radicals:
\[
\sqrt{85} \times 3\sqrt{32} = 3 \times \sqrt{85} \times \sqrt{32}
\]
Next, we can use the property of radicals that states \( \sqrt{a} \times \sqrt{b} = \sqrt{a \times b} \):
\[
3 \times \sqrt{85 \times 32}
\]
Now we compute \( 85 \times 32 \):
\[
85 \times 32 = 2720
\]
Now, we have:
\[
3 \sqrt{2720}
\]
Next, we simplify \( \sqrt{2720} \). First, we factor \( 2720 \).
Finding the prime factorization of \( 2720 \):
- \( 2720 \div 10 = 272 \)
- \( 272 = 2^4 \times 17 \) (since \( 272 = 16 \times 17 \))
So the prime factorization gives:
\[
2720 = 10 \times 272 = 10 \times 16 \times 17 = 2^4 \times 5 \times 17
\]
Now, we can rewrite \( \sqrt{2720} \):
\[
\sqrt{2720} = \sqrt{2^4 \times 5 \times 17} = \sqrt{(2^2)^2 \times 5 \times 17} = 2^2 \sqrt{5 \times 17} = 4\sqrt{85}
\]
Now substituting back, we have:
\[
3 \sqrt{2720} = 3 \times 4 \sqrt{85} = 12\sqrt{85}
\]
Thus, the product \( \sqrt{85} \times 3\sqrt{32} \) in simplest radical form is:
\[
\boxed{12\sqrt{85}}
\]
\[
\sqrt{85} \times 3\sqrt{32} = 3 \times \sqrt{85} \times \sqrt{32}
\]
Next, we can use the property of radicals that states \( \sqrt{a} \times \sqrt{b} = \sqrt{a \times b} \):
\[
3 \times \sqrt{85 \times 32}
\]
Now we compute \( 85 \times 32 \):
\[
85 \times 32 = 2720
\]
Now, we have:
\[
3 \sqrt{2720}
\]
Next, we simplify \( \sqrt{2720} \). First, we factor \( 2720 \).
Finding the prime factorization of \( 2720 \):
- \( 2720 \div 10 = 272 \)
- \( 272 = 2^4 \times 17 \) (since \( 272 = 16 \times 17 \))
So the prime factorization gives:
\[
2720 = 10 \times 272 = 10 \times 16 \times 17 = 2^4 \times 5 \times 17
\]
Now, we can rewrite \( \sqrt{2720} \):
\[
\sqrt{2720} = \sqrt{2^4 \times 5 \times 17} = \sqrt{(2^2)^2 \times 5 \times 17} = 2^2 \sqrt{5 \times 17} = 4\sqrt{85}
\]
Now substituting back, we have:
\[
3 \sqrt{2720} = 3 \times 4 \sqrt{85} = 12\sqrt{85}
\]
Thus, the product \( \sqrt{85} \times 3\sqrt{32} \) in simplest radical form is:
\[
\boxed{12\sqrt{85}}
\]
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