Question

For any second degree polynomial with two real unequal roots. The relation between Rolles point r1 and the two roots r2 is

(a) They are independent

(b) c = r1 – r2

(c) c = r1 * ^1⁄r2

(d) c = \(\frac{r_1 + r_2}{2}\)

The question was posed to me by my school teacher while I was bunking the class.

My enquiry is from Rolle’s Theorem in section Differential Calculus of Engineering Mathematics

Answer 1

Right choice is (d) c = \(\frac{r_1 + r_2}{2}\)

To explain: Given the polynomial has two real and unequal roots, we can write the polynomial in the following form

y = (x – a)(x – b)

Where a ≠ b are the two, unequal, real roots.

Now differentiation and equating to zero yields

y = (x – b) + (x -a ) = 0

Put x = c (because is the Rolles point here)

(c – a) + (c – b) = 0

2c – (a + b) = 0

c = a + b / 2 = \(\frac{r_1 + r_2}{2}\).