To find the zeroes of a quadratic polynomial f(x) = abx^2 + (b^2 - ac)x + bc, we need to use the quadratic formula. The quadratic formula is given by:

x = (-b ± sqrt(b^2 - 4ac)) / 2a

where a, b, and c are the coefficients of the quadratic polynomial.


The relationship between the zeroes of a quadratic polynomial and its coefficients is given by Vieta's formulas. Vieta's formulas state that for a quadratic polynomial ax^2 + bx + c, the sum of the zeroes is -b/a and the product of the zeroes is c/a.

To verify this relationship for the quadratic polynomial f(x) = abx^2 + (b^2 - ac)x + bc, we need to find the zeroes of the polynomial using the quadratic formula.

x = (-b ± sqrt(b^2 - 4ac)) / 2a

Substituting the coefficients of f(x), we get:

x = (- (b^2 - ac) ± sqrt((b^2 - ac)^2 - 4abcbc)) / 2ab

Simplifying the expression, we get:

x = (-b^2 + ac ± sqrt(b^4 - 2b^2ac + a^2c^2 - 4a^2bc)) / 2ab

The sum of the zeroes is given by:

(-b^2 + ac + sqrt(b^4 - 2b^2ac + a^2c^2 - 4a^2bc)) / 2ab + (-b^2 + ac - sqrt(b^4 - 2b^2ac + a^2c^2 - 4a^2bc)) / 2ab

= -b/a

The product of the zeroes is given by:

((-b^2 + ac + sqrt(b^4 - 2b^2ac + a^2c^2 - 4a^2bc)) / 2ab) * ((-b^2 + ac - sqrt(b^4 - 2b^2ac + a^2c^2 - 4a^2bc)) / 2ab)

= c/a

Therefore, we have verified the relationship between the zeroes of the quadratic polynomial f(x) = abx^2 + (b^2 - ac)x + bc and its coefficients.

Hi ,

It is given that ,

f( x ) = abx² + ( b² - ac )x - bc

To find zeroes of f( x ) , we have to

take f( x ) = 0

abx² + ( b² - ac )x - bc = 0

abx² + b² x - acx - bc = 0

bx( ax + b ) - c( ax + b ) = 0

( ax + b ) ( bx - c ) = 0

ax + b = 0 or bx - c = 0

ax = - b or bx = c

x = -b/a or x = c/b

Therefore ,

-b/a , c/b are two zeroes of f( x ).

I hope this helps you.

: )