21use product of the roots.. we can find the value of α..hence we can find the polynomial.
(a+b+c)2 is just (f(1))2, try this...
Q. If the roots of the equation ax2+bx+c=0 , are of the form αα-1 and α+1α, then the value of (a+b+c)2 is? [ans - b2-4ac]
I got the answer by putting any value of alpha but how to solve it in subjective form
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4 Answers
21
11Sum of the roots of the equaion=-ba
=αα-1+α+1α=-ba........eq(i)
Product of the roots of the eqaution=ca
=αα-1xα+1α=ca
=α+1α-1=ca
Applying componendo and dividendo..
α=c+ac-a
Putting the value in eq (i) we get,
c+a2a+2cc+a=-ba
After solving we get,
a2+c2+2ab+2bc+2ca=-4ac
Adding b2 on both sides we get,
a2+b2+c2+2ab+2bc+2ca=b2-4ac
(a+b+c)2=b2-4ac..(Hence Proved)
1thnxx khilen
i didnt knew about componendo and dividendo but now i understood it
341f(x) = a(x-r_1)(x-r_2)
Hence a+b+c = a(1-r_1)(1-r_2) = a \left(1-\frac{\alpha}{\alpha-1} \right)\left(1-\frac{\alpha+1}{\alpha} \right) = \frac{a}{\alpha(\alpha-1)}
Thus
(a+b+c)^2 = \frac{a^2}{\alpha^2(\alpha-1)^2} = a^2 \left(\frac{\alpha}{\alpha-1} - \frac{\alpha+1}{\alpha} \right)^2 = a^2 (r_1-r_2)^2
a^2[ (r_1+r_2)^2 - 4r_1r_2] = b^2-4ac