If α and β are the Zeros of the Quadratic Polynomial f(x) = ax2 + bx + c, then Evaluate

Solution:

(i) α − β

To find α – β, we need to know the values of α and β, which are the zeroes of the quadratic polynomial. Let’s assume the quadratic polynomial is f(x) = ax² + bx + c, and its zeroes are α and β.

Then we know that:

α + β = -b/a

We can rearrange this equation to get:

α = -b/a – β

Now, we substitute this expression for α into the equation f(α) = 0, which gives:

aα² + bα + c = 0

Substituting for α, we get:

a(-b/a – β)² + b(-b/a – β) + c = 0

Expanding and simplifying, we get:

aβ² + (-²ab/a)β + (a/b)c – b/a = 0

Multiplying both sides by b/a, we get:

abβ² – ²abβ + ac – b² = 0

Dividing both sides by ab, we get:

β² – (b/a)β + (c/a) = 0

This is a quadratic equation in β, with coefficients -b/a and c/a. So we can use the quadratic formula to solve for β:

β = [-(b/a) ± √((b/a)² – 4(c/a))]/²

Once we have found the values of α and β, we can easily find α – β:

α – β = (-b/a – β) – β

α – β = -b/a – 2β

α – β = -b/a – 2[(-b/a) ± √((b/a)² – 4(c/a))]/2

Simplifying, we get:

α – β = ±√(b² – 4ac)/a

(ii) 1/α − 1/β

1/α – 1/β = (β – α)/(αβ)

Using the quadratic formula, we have:

α,β = [-b ± sqrt(b² – 4ac)]/(2a)

So, substituting the expression for α and β:

1/α – 1/β = [(b + sqrt(b² – 4ac)) – (b – sqrt(b² – 4ac))]/[²a(sqrt(b² – 4ac))]

Simplifying the numerator:

1/α – 1/β = 2sqrt(b² – 4ac)/[2a(sqrt(b² – 4ac))]

1/α – 1/β = ±sqrt(b² – 4ac)/c

Therefore, the correct answer for (ii) in terms of a, b, and c is:

1/α – 1/β = ±sqrt(b² – 4ac)/c

(iii) 1/α + 1/β – 2αβ

We can first find a common denominator for the two fractions by multiplying them by (αβ)/(αβ):

1/α + 1/β – 2αβ = (β + α) / (αβ) – 2αβ

= (β + α – 2αβ²) / (αβ)

using αβ = c/a and α + β = -b/a

1/α + 1/β – 2αβ = (-b/a) / (c/a) – 2c/a

On solving we will get,

= -(b/c + 2c/a)

Therefore, 1/α + 1/β – 2αβ = -b/c – 2c/a.

(iv) α²β − αβ²

We can rewrite the expression α²β − αβ² as αβ(α − β)

On putting αβ = c/a and α − β = -b/a, we get

= c/a(− b/a)

= -cb/a²

(v) α⁴ + β⁴

Simplifying: α⁴ + β⁴ = (α² + β²)² – 2α²β²

We already know that:

α + β = -b/a

αβ = c/a

Using these values, we can find:

α² + β² = (α + β)² – 2αβ

α² + β² = (b/a)² – 2(c/a)

α² + β² = (b² – 2ac)/a²

Substituting this back into the original identity, we get:

α⁴ + β⁴ = ((b² – ²ac)/a²)² – 2α²β²

α⁴ + β⁴ = (b⁴ – 4b²ac + 4a²c²)/a⁴ – 2(c/a)²

α⁴ + β⁴ = (b⁴ – 4b²ac + 4a²c² – 2c²a²)/a⁴

Simplifying this expression, we get:

α⁴ + β⁴ = (b⁴ – 4b²ac + 4a²c² – 2c²a²)/a⁴

α⁴ + β⁴ = (b⁴ – 4b²ac + 4c²(a² – c))/a⁴

Therefore, the value of α⁴ + β⁴ in terms of a, b, and c is:

α⁴ + β⁴ = (b⁴ – 4b²ac + 4c²(a² – c))/a⁴

or

α⁴ + β⁴ = (b² – 2ac)²– 2a²c²))/a⁴

If α and β are the Zeros of the Quadratic Polynomial f(x) = ax2 + bx + c, then Evaluation Answers

(i) α − β = ±sqrt(b² – 4ac)/a

(ii) 1/α − 1/β = ±sqrt(b² – 4ac)/c

(iii) 1/α + 1/β – 2αβ =  -b/c – 2c/a

(iv) α²β − αβ² = -cb/a²

(v) α⁴ + β⁴ = (b² – 2ac)²– 2a²c²))/a⁴

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