Question

Use the Pythagorean identity with X=8 and Y=5 to generate a Pythagorean triple

Answer 1

`\huge\text{Hey there!}`

`\huge\textbf{Assuming the formula:}`

`\mathbf{a^2 + b^2 = c^2}`

`\huge\textbf{If so:}`

`\mathbf{x^2 + y^2 = c^2}`

`\mathbf{8^2 + 5^2 = c^2}`

`\mathbf{8* 8 + 5 * 5 = c^2}`

`\mathbf{64 + 25 = c^2}`

`\mathbf{89 = c^2}`

`\huge\textbf{Subtract \boxed{\rm{\bf c^2}} to both sides:}`

`\mathbf{89 - c^2 = c^2 - c^2}`

`\huge\textbf{Simplify it: }`

`\mathbf{-c^2 + 89 = 0}`

`\huge\textbf{Subtract \boxed{\bf 89} to both sides:}`

`\mathbf{-c^2 + 89 - 89 = 0 - 89}`

`\huge\textbf{Simplify it:}`

`\mathbf{-c^2 = -89}`

`\huge\textbf{Divide \boxed{\bf{-1}} to both sides:}`

`\mathbf{(-c^2)/(-1) = (-89)/(-1)}`

`\huge\textbf{Simplify it:}`

`\mathbf{-c^2 = -89}`

`\huge\textbf{Take the square root of the number:}`

`\mathbf{c = \pm √(89)}`

`\huge\textbf{Thus, this possibly means:}`

`\mathbf{c = √(89) \ or\ c = -√(89)}`

`\huge\textbf{Thus, your answer possibly is:}`

`\huge\boxed{\mathsf{c = √(89) \ or\ c = -√(89)}}\huge\checkmark`

~
`\frak{Amphitrite1040:)}`