MCQ
Write a pythagorean triplet whose one member is $17.$
- A$17, 12, 13$
- ✓$8, 15, 17$
- C$17, 24, 32$
- D$16, 17, 34$
✓
Answer
Correct option: B.
$8, 15, 17$
If we take $x^2 + 1 = 17$
So, $x^2 = 16$
$\Rightarrow$ $x = 4$
Therefore, $2x = 8$ and $x^2 - 1 = (4)^2 - 1 = 16 - 1 = 15$
Therefore, the required triplet is $8, 15$ and $17.$
So, $x^2 = 16$
$\Rightarrow$ $x = 4$
Therefore, $2x = 8$ and $x^2 - 1 = (4)^2 - 1 = 16 - 1 = 15$
Therefore, the required triplet is $8, 15$ and $17.$
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