Question

If 'a' and 'B' are zeroes of the quadratic polynomial ax + bx + c, a O show that:
a +B= -b/a and a ß= c/a,​

Answer 1

A + B = $\frac{- b}{a}$ and AB = $\frac{c}{a}$

Step-by-step explanation:

Let A and B be the zeros of a quadratic polynomial.

Let P(x) = a$x^{2}$ + bx + c, when a >0      ..... (1)

Then, (x - A) and (x - B) are the factors of P(x).

∴  a$x^{2}$ + bx + c = k(x - A)(x - B), where k is constant

= k($x^{2}$  - (A + B)x + AB ),

= k$x^{2}$ - k(A + B)x + kAB          ..... (2)

On comparing coefficients of like powers x on both sides in equations (1) and (2), we get

k = a , - k(A + B) = b and k(AB) = c

Put k = a, we get

∴ A + B = $\frac{- b}{a}$ and AB = $\frac{c}{a}$

Hence, it is proved.