Quadratic equations question somebody pls answer step by step
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Let P be the pole to be erected and A, B be the opposite fixed gates.
Given, PA – PB = 7
Let PA = a, PB = b
Hence a – b = 7
⇒ a = 7 + b ...(1)
In right ΔPAB,
AB² = AP² + BP² (By Pythagoras theorem)
17² = a² + b²
289 = (7 + b)² + b²
289 = 49 + 14b + 2b²
2b² + 14b – 240 = 0
b² + 7b – 120 = 0
b² - 8b +15b – 120 = 0
b(b + 8) - 15(b + 8) = 0
(b + 8)(b – 15) = 0
Hence b = 15 or – 8
But ‘b’ cannot be negative
Therefore, b = 15
⇒ a = 7 + 15 = 22
Hence, PA = 22 m and PB = 15 m
Given, PA – PB = 7
Let PA = a, PB = b
Hence a – b = 7
⇒ a = 7 + b ...(1)
In right ΔPAB,
AB² = AP² + BP² (By Pythagoras theorem)
17² = a² + b²
289 = (7 + b)² + b²
289 = 49 + 14b + 2b²
2b² + 14b – 240 = 0
b² + 7b – 120 = 0
b² - 8b +15b – 120 = 0
b(b + 8) - 15(b + 8) = 0
(b + 8)(b – 15) = 0
Hence b = 15 or – 8
But ‘b’ cannot be negative
Therefore, b = 15
⇒ a = 7 + 15 = 22
Hence, PA = 22 m and PB = 15 m
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