Find real parameter p such that the solutions to the equation (p - 3).x² + (p²+1)x - 11p+18 = 0 are lengths of the legs of the right triangle with the hypotenuse length √17
Answers
Given : solutions to the equation (p - 3).x² + (p²+1)x - 11p+18 = 0 are lengths of the legs of the right triangle with the hypotenuse length √17
To Find : Real values of p
Solution:
(p - 3).x² + (p²+1)x - 11p+18 = 0
α + β = - (p²+1) / (p - 3)
αβ =(18 - 11p) / (p - 3)
α² + β² = 17 ( lengths of the legs of the right triangle with the hypotenuse length √17 )
α + β = - (p²+1) / (p - 3)
Squaring both sides
=> α² + β² + 2αβ = (p²+1)² / (p - 3)²
=> 17 + 2(18 - 11p) / (p - 3) = (p²+1)² / (p - 3)²
=> 17 (p - 3)² + 2(18 - 11p) (p - 3) = (p²+1)²
=> 17(p² - 6p + 9) + 2(-11p² + 51p - 54) = p⁴ + 2p² + 1
=> 17 p² - 102p + 153 -22p² +102p - 108 = p⁴ + 2p² + 1
=> -5p² + 45 = p⁴ + 2p² + 1
=> p⁴ + 7p² - 44 = 0
=> p⁴ + 11p² - 4p² - 44 = 0
=> p²(p² + 11) - 4(p² + 11) = 0
=> (p² - 4)(p² + 11) = 0
=> p² = 4 , p² = - 11 ( imaginary )
=> p = ± 2
p = ± 2
Learn More:
Values of k for which the quadratic equation 2x^2 – kx + k = 0 has ...
brainly.in/question/8043948
What kind of roots or solution each of the quadratic equation? b ...
brainly.in/question/25658971