Question

find the roots of the quadratic equation where p is constant x^2+x-p(p+1)=0​

Answer 1

x^2 + x - p^2 -p =0

=> x^2- p^2 + (x-p)=0

=> (x+p)(x-p)+(x-p) =0

=> (x-p)(x+p+1)=0

=> (x-p) = 0. ; (x+p+1) =0

=> x= p ; x= -(p+1)

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Answer 2

Introduction on the problem :

A root is a solution of the given equation .

It is the number that satisfies the value of the solution .

For example x² + 2 x + 1 = f ( x ) is a function .

Then the value which satisfies the value of f ( x ) is a root .

Given :

x² + x - p ( p + 1 ) = 0

By splitting method :

⇒ x² + x - p ( p + 1 )

We will split x ( 1 ) into x ( p - p + 1 )

⇒ x² + x ( p - p + 1 ) - p ( p + 1 )

⇒ x² - p x + x ( p + 1 ) - p ( p + 1 )

⇒ x ( x - p ) + ( p + 1 )( x - p )

⇒ ( x + p + 1 )( x - p ) = 0

Either :

x + p + 1 = 0

⇒ x = - ( p + 1 )

Or :

x - p = 0

⇒ x = p

ANSWER :

The 2 roots are p and - ( p + 1 ) .