34. If a and B are the zeros of the quadratic polynomial p(x) = x2 – 2x + 3, find a polynomial
whose zeros are a + 2 and B +

2 thoughts on “34. If a and B are the zeros of the quadratic polynomial p(x) = x2 – 2x + 3, find a polynomial<br />whose zeros are a + 2 and B +”

1. Solution :

   Given Equation

   [tex]\bf \red{x^{2} – 2x + 3}[/tex]

   Sum of the zeroes

   Zeroes = 1 and 2

   [tex]\mathbb Sum = \dfrac{-(-2)}{1}[/tex]

   [tex]\sf Sum = \dfrac{2}{1}[/tex]

   Sum = 2

   Product of zeroes

   [tex]\alpha \beta = \dfrac{3}{1}[/tex]

   [tex]\alpha \beta = 3[/tex]

   When added by 2

   [tex]\bf \alpha + 2+ \beta + 2[/tex]

   [tex]\mathbb \alpha + \beta + 4[/tex]

   According to question (A/q)

   [tex]\alpha + \beta = 2[/tex]

   [tex]2 + \beta = 6[/tex]

   [tex]\sf \beta = 4[/tex]

   In product:

   [tex]\bigg(\alpha + 2\bigg) ,\bigg(\beta + 2\bigg)[/tex]

   [tex]\sf \alpha \beta + 2\alpha + 2\beta + 4[/tex]

   [Taking 2 as common]

   [tex]\sf \alpha \beta + 2(\alpha + \beta ) + 4[/tex]

   [tex]\sf 3 + 2(2) + 4[/tex]

   [tex]\sf 3 + 4 + 4[/tex]

   11

   Therefore the Equation be formed :-

   x² – 6x + 11

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2. Solution :

   Given Equation

   [tex]\bf \red{x^{2} – 2x + 3}[/tex]

   Sum of the zeroes

   Zeroes = 1 and 2

   [tex]\sf Sum = \dfrac{-(-2)}{1}[/tex]

   [tex]\sf Sum = \dfrac{2}{1}[/tex]

   Sum = 2

   Product of zeroes

   [tex]\alpha \beta = \dfrac{3}{1}[/tex]

   [tex]\alpha \beta = 3[/tex]

   When added by 2

   [tex]\bf \alpha + 2+ \beta + 2[/tex]

   [tex]\sf \alpha + \beta + 4[/tex]

   According to the question

   [tex]\alpha + \beta = 2[/tex]

   [tex]2 + \beta = 6[/tex]

   [tex]\sf \beta = 4[/tex]

   In product

   [tex]\bigg(\alpha + 2\bigg) ,\bigg(\beta + 2\bigg)[/tex]

   [tex]\sf \alpha \beta + 2\alpha + 2\beta + 4[/tex]

   Taking 2 as common

   [tex]\sf \alpha \beta + 2(\alpha + \beta ) + 4[/tex]

   [tex]\sf 3 + 2(2) + 4[/tex]

   [tex]\sf 3 + 4 + 4[/tex]

   [tex]11[/tex]

   Equation formed :-

   x² – 6x + 11

   Reply