Question

If alpha and beta are the zeroes of the quadratic polynomial p(x)=ax2+bx+c then evaluate α-β

Answer 1

Step-by-step explanation:

p(x) = ax² + bx + c

zeroes are Alpha & Beta of a above quadratic polynomial .

alpha + beta = -b/a

alpha*beta = c/a

---------------------------

alpha - Beta = √(alpha+beta)²-4alpha*beta

= √(b²/a²) -4c/a

= (√b²-4ac) / a = √D/a

from Shridharacharya formula :- (b²-4ac) = D

Answer 2

Answer:

√{ [ (ab² - 4c³)/c²a ] }

Step-by-step explanation:

Given that α and β are the zeroes of the polynomial p(x) = ax² + bx + c.

On comparing this with ax² + bx + c, we get

a = a, b = b, c = c

Now,

• Sum of zeroes = α + β = - b/c

→ - b/c

• Product of zeroes = αβ = c/a

→ c/a

Now,

• Squaring both sides of sum of zeroes

→ (α + β)² = (- b/c)²

Identity : (a + b)² = a² + b² + 2ab

Here, a = α, b = β

→ α² + β² + 2αβ = b²/c²

→ α² + β² + 2(c/a) = b²/c²

→ α² + β² + (2c/a) = b²/c²

→ α² + β² = (b²/c²) - (2c/a)

→ α² + β² = (ab² - 2c³)/c²a

Now,

• Squaring of α - β

→ (α - β)²

Identity : (a - b)² = a² + b² - 2ab

Here, a = α, b = β

→ α² + β² - 2αβ

→ [ (ab² - 2c³)/c²a ] - 2(c/a)

→ [ (ab² - 2c³)/c²a ] - (2c/a)

→ [ (ab² - 2c³ - 2c³)/c²a ]

→ [ (ab² - 4c³)/c²a ]

We have

→ (α - β)² = [ (ab² - 4c³)/c²a ]

→ α - β = √{ [ (ab² - 4c³)/c²a ] }