Question
use the polynomial identity (x^2+x^y)^2= (x^2-y^2)^2+(2xy)^2 to generate a pythagorean triple when x equals 12 and y equals 5 which of the following is one of the values in the Pythagorean triple
A. 49
B.60
C.289
D.119
A. 49
B.60
C.289
D.119
Answers
Answer
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Answered by
GPT-4o mini
To generate a Pythagorean triple using the given polynomial identity \((x^2 + xy)^2 = (x^2 - y^2)^2 + (2xy)^2\), we'll substitute \(x = 12\) and \(y = 5\).
First, we calculate \(x^2\) and \(xy\):
1. \(x^2 = 12^2 = 144\)
2. \(xy = 12 \cdot 5 = 60\)
Now, substituting these values into the identity:
1. Compute \(x^2 - y^2\):
\[
x^2 - y^2 = 144 - 25 = 119
\]
2. Compute \(2xy\):
\[
2xy = 2 \cdot 60 = 120
\]
Now, we can now identify the sides of the resulting Pythagorean triple using the following:
- One leg: \(x^2 - y^2 = 119\)
- The other leg: \(2xy = 120\)
- The hypotenuse can be found using the left hand side:
\[
(x^2 + xy)^2 = (144 + 60)^2 = 204^2 = 41616
\]
The hypotenuse is:
\[
\sqrt{(x^2 + xy)^2} = 204
\]
The Pythagorean triple is thus \( (119, 120, 204) \).
Among the choices provided:
A. 49
B. 60
C. 289
D. 119
The value from the Pythagorean triple we found is \(119\).
**So, the answer is D. 119.**
First, we calculate \(x^2\) and \(xy\):
1. \(x^2 = 12^2 = 144\)
2. \(xy = 12 \cdot 5 = 60\)
Now, substituting these values into the identity:
1. Compute \(x^2 - y^2\):
\[
x^2 - y^2 = 144 - 25 = 119
\]
2. Compute \(2xy\):
\[
2xy = 2 \cdot 60 = 120
\]
Now, we can now identify the sides of the resulting Pythagorean triple using the following:
- One leg: \(x^2 - y^2 = 119\)
- The other leg: \(2xy = 120\)
- The hypotenuse can be found using the left hand side:
\[
(x^2 + xy)^2 = (144 + 60)^2 = 204^2 = 41616
\]
The hypotenuse is:
\[
\sqrt{(x^2 + xy)^2} = 204
\]
The Pythagorean triple is thus \( (119, 120, 204) \).
Among the choices provided:
A. 49
B. 60
C. 289
D. 119
The value from the Pythagorean triple we found is \(119\).
**So, the answer is D. 119.**
Answered by
GPT-4o mini
Let’s double-check the calculations step by step to confirm the Pythagorean triple generated by the polynomial identity \((x^2 + xy)^2 = (x^2 - y^2)^2 + (2xy)^2\) with \(x = 12\) and \(y = 5\).
1. **Calculate \(x^2\) and \(y^2\)**:
- \(x^2 = 12^2 = 144\)
- \(y^2 = 5^2 = 25\)
2. **Calculate \(xy\)**:
- \(xy = 12 \cdot 5 = 60\)
3. **Calculate \(x^2 - y^2\)**:
- \(x^2 - y^2 = 144 - 25 = 119\)
4. **Calculate \(2xy\)**:
- \(2xy = 2 \cdot 60 = 120\)
5. **Calculate \(x^2 + xy\)**:
- \(x^2 + xy = 144 + 60 = 204\)
Now we summarize the key values obtained:
- One leg (from \(x^2 - y^2\)): **119**
- The other leg (from \(2xy\)): **120**
- The hypotenuse (from \(x^2 + xy\)): **204**
Thus, the Pythagorean triple formed is \((119, 120, 204)\).
Regarding the choices presented:
A. 49
B. 60
C. 289
D. 119
Based on the validation, **D. 119 is indeed a value in the Pythagorean triple.**
1. **Calculate \(x^2\) and \(y^2\)**:
- \(x^2 = 12^2 = 144\)
- \(y^2 = 5^2 = 25\)
2. **Calculate \(xy\)**:
- \(xy = 12 \cdot 5 = 60\)
3. **Calculate \(x^2 - y^2\)**:
- \(x^2 - y^2 = 144 - 25 = 119\)
4. **Calculate \(2xy\)**:
- \(2xy = 2 \cdot 60 = 120\)
5. **Calculate \(x^2 + xy\)**:
- \(x^2 + xy = 144 + 60 = 204\)
Now we summarize the key values obtained:
- One leg (from \(x^2 - y^2\)): **119**
- The other leg (from \(2xy\)): **120**
- The hypotenuse (from \(x^2 + xy\)): **204**
Thus, the Pythagorean triple formed is \((119, 120, 204)\).
Regarding the choices presented:
A. 49
B. 60
C. 289
D. 119
Based on the validation, **D. 119 is indeed a value in the Pythagorean triple.**
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