Latest m4maths Aptitude Question SOLUTION: If a,b are the roots of x^2+px+q=0 then x^2+(a+b-ab)x-ab(a+b)=0 has roots

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If a,b are the roots of x^2+px+q=0 then x^2+(a+b-ab)x-ab(a+b)=0 has roots

Asked In m4maths Satyaranjan (11 years ago)

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Devendra Marghade  (11 years ago)

roots= -p,-q
As for x^2+px+q=0 ---(i), roots are a & b, then a+b=-p, ab=q
Eqn x^2+(a+b-ab)x-ab(a+b)=0 ---(ii)
Substituting values of a+b & ab in (ii),
we get eqn.(ii)as, x^2-(p+q)x+pq=0, so the roots are -p,-q

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Satyaranjan  (11 years ago)

Options are1) -p,-q
2) p,-q
3) a,b
4) -a,-b
5) a,-b
6) p,q
7) -p,q
8) -a,b
9) ap,bq
10)None of these

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Satyaranjan  (11 years ago)

If a and b are roots then a+b=-p and ab=q,So the equation x^2+(a+b-ab)x-ab(a+b)=0 becomes x^2+(-p-q)x+pq=0.Again let it has roots b and c,then b+c=-(-p-q)=p+q and bc=pq,then (b-c)^2=(b+c)^2-4bc=(p+q)^2-4pq=(p-q)^2,so b-c=p-q,sop we get b=p and c=q,So the roots are p and q.

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Satyaranjan  (11 years ago)

@Devendra M
Your answer is -p and -q( x^2-(p+q)x+pq=0, so the roots are -p,-q),that means these must satisfy the equation x^2-(p+q)x+pq=0,but these do not .You can check.But if you take p and q as your roots the equation becomes 0 for both the roots.So the answer is p and q.

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