If α and β are the Zeroes of the Quadratic Polynomial p(x) = x^2– x – 2, Find a Polynomial Whose Zeroes are 2α + 1 and 2β + 1

If α and β are the Zeroes of the Quadratic Polynomial p(x) = x²– x – 2, Find a Polynomial Whose Zeroes are 2α + 1 and 2β + 1

Solution:

Given that α and β are the zeroes of the quadratic polynomial p(x) = x²- x - 2, we can find the sum and product of the zeroes using the Vieta's formulas.

The sum of the zeroes is α + β = -(-1)/1 = 1, and the product of the zeroes is αβ = -2/1 = -2.

Now, let's find the polynomial with zeroes 2α + 1 and 2β + 1.

The sum of the new zeroes is (2α + 1) + (2β + 1) = 2(α + β) + 2 = 2(1) + 2 = 4.

The product of the new zeroes is (2α + 1)(2β + 1) = 2(αβ) + 2(α) + 2(β) + 1 = 2(-2) + 2(α) + 2(β) + 1 = -4 + 2α + 2β + 1 = 2α + 2β - 3.

Therefore, the polynomial with zeroes 2α + 1 and 2β + 1 is q(x) = x² - (sum of the zeroes)x + product of the zeroes.

Substituting the values, we have q(x) = x²- 4x - 5.

So, the polynomial with zeroes 2α + 1 and 2β + 1 is indeed q(x) = x² - 4x - 5.

Answer:

If α and β are the Zeroes of the Quadratic Polynomial p(x) = x²– x – 2, then a Polynomial Whose Zeroes are 2α + 1 and 2β + 1 is q(x) = x^2 - 4x - 5

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