Question

if the roots of a quadratic equation(a-b)$x^{2}$+(b-c)x+(c-a)=oare equal,prove that 2a=b+c

Answer 1

$(a - b)x^2 + (b - c)x + (c - a) = 0$

Given: roots are equal.

Therefore, the discriminant of this quadratic equation must be zero.

Or, $D = B^2 - 4AC = 0$ (for a quadratic $Ax^2 + Bx + C = 0$)

So, substituting the values of B, A and C, on comparing with standard quadratic, we get:

$D = (b - c)^2 - 4(a - b)(c - a) = 0$

Or, $b^2 + c^2 - 2bc - 4ac + 4bc + 4a^2 - 4ab = 0$

Or, $b^2 + c^2 + 4a^2 + 2(bc - 2ac - 2ab) = 0$

The above, through observation*, can be written as:

$(b + c - 2a)^2 = 0$

=> $b + c - 2a = 0$

Or, $b + c = 2a.$

Hence proved.

* If you aren't able to observe that, then with a little extra work, you could still simplify it:

$b^2 + c^2 + 2bc + 4a^2 - 4ab - 4ac = 0$

$(b + c)^2 - 4a(b + c) + 4a^2 = 0$

This must again be solved through observation. This one is a little more noticeable though.

$(b + c)^2 + (2a)^2 - 2.2a.(b + c) = 0$

Comparing with $(a - b)^2 = a^2 + b^2 - 2ab$, we can write it as:

$(b + c - 2a)^2 = 0$

And then the rest of the steps as above.

Observation, in this question, is necessary.