2 answers:
Use the Pythagorean Theorem of a² + b² = c² to solve for x.
x² + (x + 3)² = (√117)²
x² + x² + 6x + 9 = 117
2x² + 6x + 9 = 117
2x² + 6x + 9 - 117 = 0
2x² + 6x - 108 = 0
2x² + 18x - 12x - 108 = 0
2x(x + 9) - 12(x + 9) = 0
(2x - 12)(x + 9) = 0
x = - 9, 6
Length cannot be negative so you can't use - 9.
x = 6. Option C is your answer.
Use the Pythagorean Theorem:
a² + b² = c²
(x + 3)² + (x)² = (√117)²
(x² + 6x + 9) + (x²) = 117
2x² + 6x + 9 = 117
2x² + 6x - 108 = 0
2(x² + 3x - 54) = 0
2(x + 9)(x - 6) = 0
2≠0, x + 9 = 0, x - 6 = 0
x = - 9, x = 6
Length cannot be negative so x = -9 is not valid.
Answer: x = 6
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