Question

Please help me with this.

Answer 1

Step-by-step explanation:

Given -

f(x) = x²/a + b/ac × -x/b - 1/c = 0

To Find -

Zeroes of the polynomial

Now,

x²/a + b/ac × -x/b - 1/c = 0

» x²/a - bx/abc - 1/c = 0

» bcx² - bx - ab/abc = 0

» bcx² - bx - ab = 0

here,

a = bc

b = -b

c = -ab

Using Quadratic formula :-

• x = -b ± √b² - 4ac/2a

» -(-b) ± √(-b)² - 4×bc×-ab/2(bc)

» b ± √b² + 4ab²c/2bc

Zeroes are -

x = b + √b² + 4ab²c/2bc

and

x = b - √b² + 4ab²c/2bc

Verification :-

• α + β = -b/a

» b + √b² + 4ab²c/2bc + b - √b² + 4ab²c/2bc = -(-b)/bc

» b + √b² + 4ab²c + b - √b² + 4ab²c/2bc = b/bc

» 2b/2bc = 1/c

• » 1/c = 1/c

LHS = RHS

And

• αβ = c/a

» b + √b² + 4ab²c/2bc × b - √b² + 4ab²c/2bc= -ab/bc

» b² - (b² + 4ab²c)/4b²c²= -a/c

» b² - b² - 4ab²c/4b²c² = -a/c

» -4ab²c/4b²c² = -a/c

• » -a/c = -a/c

LHS = RHS

Hence,

Verified..