1.The sum of the roots of the equation x+a1+x+b1=c1 is zero.Prove that the product of the roots is −21(a2+b2).
2.If the roots of the equation p(q−r)x2+q(r−p)x+r(p−q)=0 be equal , then show that p1+r1=q2.
3.Find the condition that one root of ax2+bx+c=0 shall be ′n′ times the other.
4.If Sinx+Siny=a and Cosx+Cosy=b,Then find the roots of equation (a2+b2+2b)t2−4at+a2+b2−2b=0
#Algebra
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Simplified expression is,
⇒x2+x(a+b−2c)+(ab−ac−bc)=0.
Using vieta's formula,
Sum of roots = 2c−a−b=0.
Product of roots = ab−ac−bc.
=ab−a(2a+b)−b(2a+b)
=−21(a2+b2)
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Thank u very much. :)
A - 3. Let the one root of given equation is β, then the other root must be nβ.
Using Vieta's formula, Sum of roots = β+nβ=−ab⇒β=−a(1+n)b.
Product of roots = nβ2=ac.
⇒n(a(1+n)−b)2=ac.
⇒nb2=ac+acn2+2anc.
⇒acn2+(2ac−b2)n+ac=0.
For real value of n, discriminant ≥ 0.
⇒(2ac−b2)2−4(ac)2≥0.
⇒b≥2ac
For the second one, you get the root as 1 by observation and then just apply Vieta's and get the answer
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Can u please show it.
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Sum of roots=2=pq−rppq−rq》》2pq−2rp=pq−rq》》pq+rq=2rp Divide by pqr to get required result