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
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...
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