If a+b+c=0 and a,b,c are real, then prove that the eqn
(b-x)2 - 4 (a-x)(c-x) = 0 has real roots and the roots will not be equal unless a=b=c.
1MY WORKING
Since a,b,c are real, then the eqn
(b-x)2 - 4 (a-x)(c-x) = 0 has real roots if D>0,
that is (b2+x2-2bx) - 4 (ac-ax-cx+x2)=0
b2+x2-2bx - 4ac+4ax+4cx-4x2=0
-3x2 +2x(2a+2c-b) +b2-4ac=0
Now since D>0 [2(2a+2c-b)] 2 +4(3) (b2-4ac) > 0
(2a+2c-b)2 + 3(b2-4ac)>0
But the roots are real equal only if D=0 If we put a=b=c, the roots r equal and D is coming to zeo.
But I am not able to prove D>0 or roots are real???
341If a+b+c = 0, then ax2+bx+c = 0 has real roots and hence we must have b2≥4ac.
Let f(x) = (b-x)2-4(a-x)(c-x)
Case 1: If b2=4ac, then this means both roots of ax2+bx+c = 0 are 1 and hence c=a and b = -2a.
So the second quadratic is (x+2a)2-4(x-a)2 which has roots x = 0 and x = 4a.
If a = 0, then both roots are zero. But a = 0, means b =c = 0 in this case.
Case 2: If b2>4ac, then we see that f(0) >0 but, as x→∞, f(x) → - ∞ and also as x→-∞, f(x) → - ∞.
This tells us that f(x) has two real roots one positive and the other negative and they obviously cannot be equal