Consider (using capital letters to distinguish the coefficients from the given problem).
If one applies the quadratic formula, they get the roots:
The roots are real and repeated when or we can say when (this is called the discriminant).
Compare A,B,C to the given problem:
A = b-c
B = c-a
C = a-b
Now use the right hand sides of these to get the discriminant:
This can be expanded, then simplified:
divide thru by 4:
Notice this is a perfect square:
Which has the solution:
Recall:
It is this relationship between a,b,and c that must hold in order for the original descriminant to be zero. That in turn, implies repeated real roots.
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The solution by the tutor @math_helper is fine and correct.
In my post, I want to show you much shorter and more geometric solution to this problem.
Notice that (b-c) + (c-a) + (a-b) = 0. (It is obvious !)
It means that x= 1 is the root to this quadratic.
But the problem states that the root is REPEATED (!)
It means that the quadratic has the minimum (or the maximum) at the point x= 1.
In any case, it means that the point (1,0) in the coordinate plane is the VERTEX of the quadratic function/(parabola).
It implies that
1 = -, (1) (well known formula for the parabola's symmetry axis)
where "A" is the coefficient at x^2 and "B" is the coefficient at x.
In our case A = b-c and B = c-a; hence, (1) means that
1 = -, or
2*(b-c) = -(c-a),
2b - 2c = -c + a
2b = -c + a + 2c = a + c
b =
I hope that after reading my solution, you will better understand, why this statement takes place.