The root of the constant x2+x-p(p+1)=0 where p is a constsant
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Answered by
30
The answer is given below :
Given,
x² + x - p(p + 1) = 0
=> x² + x - p² - p = 0
=> x² - p² + x - p = 0
=> (x + p)(x - p) + (x - p) = 0
=> (x - p)(x + p + 1) = 0
Either, x - p = 0 or, x + p + 1 = 0
=> x = p and x = - p - 1
Thus the required roots of the given equation are
x = p, - p - 1.
Thank you for your question.
Given,
x² + x - p(p + 1) = 0
=> x² + x - p² - p = 0
=> x² - p² + x - p = 0
=> (x + p)(x - p) + (x - p) = 0
=> (x - p)(x + p + 1) = 0
Either, x - p = 0 or, x + p + 1 = 0
=> x = p and x = - p - 1
Thus the required roots of the given equation are
x = p, - p - 1.
Thank you for your question.
Answered by
8
X=-p-1,p
Hope it helps you
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