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(#M40035076) OTHERS Numerical Ability question Keep an EYE

If roots of equation x^2-bx+c=0 be two consecutive integers, then b^2-4c equals?

Asked In Others harshal agrawal (11 years ago)

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steffy edmond  (11 years ago)

let the roots of the given equation be x and (x+1)=>two consecutive integers
then sum of roots= (x+(x+1))=-b/a,(a=1 in the given equation)
so (2x+1)=-b
(-2x-1)=b => (1)
product of roots= (x(x+1)=c/a,(a=1 in the given equation)
so (x^2+x)=c => (2)
therefore substituting the values of b and c in b^2-4c we get
(-2x-1)^2-4(x^2+x)= 4x^2+4x+1-4x^2-4x = 1
so b^-4c= 1

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shree  (11 years ago)

if b=3 and c=2 then x^2-3x+2=0
x^2-2x-x+2=0
x(x-2)-1(x-2)=0
x=1,2
b^2-4c= 3^2-(4*2)=9-8=1

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Divya  (11 years ago)

x^2-bx+c===>two integer
b^2-4c=====>two integers

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