If alpha and beta are zeroes if the polynomial 2x²-7x+5 then find a polynomial whose zeroes are 2 alpha + 1 and 2 beta+ 3 ??
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If α and β are the zeros of the polynomial 2x² - 7x + 5, then by Vieta's formulas, their sum is (-b)/a and their product is c/a.
Given: α + β = 7/2 αβ = 5/2
Now, if the zeros of a polynomial are (2α + 1) and (2β + 3), then by using the sum and product of roots, we can find the polynomial.
The sum of the new roots: (2α + 1) + (2β + 3) = 2(α + β) + 4
The product of the new roots: (2α + 1)(2β + 3) = 2(αβ) + 6(α + β) + 3
Now, we can use these to form the polynomial.
The polynomial is: [x - (2α + 1)][x - (2β + 3)] = 0
Expanding this expression will give the polynomial.
Polynomial: (x - (2α + 1))(x - (2β + 3)) = 0
Expanding: x² - (2α + 2β + 4)x + (2α + 1)(2β + 3)
Now, we replace (2α + 2β + 4) with 7 (from the sum of roots) and (2α + 1)(2β + 3) with 5/2 (from the product of roots):
x² - 7x + 5/2
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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