Question

If the roots of the equation x² + px + q = 0 differ by 1, then
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Answer 1

Answer:

q = p( p+1)

Step-by-step explanation:

let one root= alfa

another root = alfa +1

=> alfa+(alfa+1) = -b/a

=> 2alfa = -(a+b) / a

=> alfa = -(a+b) /2a...............(1)

again

alfa×(alfa+1) = c/a

=> alfa²+alfa = c/a

=> (a+b)²/a² -(a+b)/a = c/a

=> (a+b)²-a(a+b) = ac

=> a²+b²+2ab-a²-ab = ac

=> b(a+b) = ac

=> c = (b/a) ( a+b)

comparing the coefficients

a= 1 , b= p , c = q

=> q = p(p+1)

this is the relationship between the coefficients

of the given quadratic equation.