Question

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.​

Answer 1

Solution :

Given Equation

$\bf \red{x^{2} - 2x + 3}$

Sum of the zeroes

Zeroes = 1 and 2

$\sf Sum = \dfrac{-(-2)}{1}$

$\sf Sum = \dfrac{2}{1}$

Sum = 2

Product of zeroes

$\alpha \beta = \dfrac{3}{1}$

$\alpha \beta = 3$

When added by 2

$\bf \alpha + 2+ \beta + 2$

$\sf \alpha + \beta + 4$

According to the question

$\alpha + \beta = 2$

$2 + \beta = 6$

$\sf \beta = 4$

In product

$\bigg(\alpha + 2\bigg) ,\bigg(\beta + 2\bigg)$

$\sf \alpha \beta + 2\alpha + 2\beta + 4$

Taking 2 as common

$\sf \alpha \beta + 2(\alpha + \beta ) + 4$

$\sf 3 + 2(2) + 4$

$\sf 3 + 4 + 4$

$11$

Equation formed :-

x² - 6x + 11

Answer 2

Solution :

Given Equation

$\bf \red{x^{2} - 2x + 3}$

Sum of the zeroes

Zeroes = 1 and 2

$\mathbb Sum = \dfrac{-(-2)}{1}$

$\sf Sum = \dfrac{2}{1}$

Sum = 2

Product of zeroes

$\alpha \beta = \dfrac{3}{1}$

$\alpha \beta = 3$

When added by 2

$\bf \alpha + 2+ \beta + 2$

$\mathbb \alpha + \beta + 4$

According to question (A/q)

$\alpha + \beta = 2$

$2 + \beta = 6$

$\sf \beta = 4$

In product:

$\bigg(\alpha + 2\bigg) ,\bigg(\beta + 2\bigg)$

$\sf \alpha \beta + 2\alpha + 2\beta + 4$

[Taking 2 as common]

$\sf \alpha \beta + 2(\alpha + \beta ) + 4$

$\sf 3 + 2(2) + 4$

$\sf 3 + 4 + 4$

11

Therefore the Equation be formed :-

x² - 6x + 11