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

EXPLANATION.

α,β are the zeroes of the quadratic equation.

⇒ x² - 2x + 3 = 0.

As we know that,

Sum of the zeroes of the quadratic equation.

⇒ α + β = -b/a.

⇒ α + β = -(-2)/1 = 2.

Products of the zeroes of the quadratic equation.

⇒ αβ = c/a.

⇒ αβ = 3/1 = 3.

Zeroes of the quadratic polynomial = α + 2 and β + 2.

As we know that,

Sum of zeroes of the quadratic equation.

⇒ α + β.

⇒ α + 2 + β + 2.

⇒ α + β + 4.

Put the values of α + β = 2 in equation, we get.

⇒ 2 + 4 = 6.

Products of the zeroes of the quadratic equation.

⇒ αβ.

⇒ (α + 2)(β + 2).

⇒ α(β + 2) + 2(β + 2).

⇒ αβ + 2α + 2β + 4.

⇒ αβ + 2[α + β] + 4.

Put the value of α + β = 2 & αβ = 3 in equation, we get.

⇒ 3 + 2[2] + 4.

⇒ 3 + 4 + 4 = 11.

As we know that,

Formula of a Quadratic polynomial.

⇒ x² - (α + β)x + αβ.

Put the values in the equation, we get.

⇒ x² - (6)x + 11 = 0.

⇒ x² - 6x + 11 = 0.

                                                                                                                         

MORE INFORMATION.

Maximum and minimum value of quadratic expression.

In a quadratic expression ax² + bx + c.

(1) = If a > 0, quadratic expression has least value at x = -b/2a. This least value is given by 4ac - b²/4a = -D/4a.

(2) = If a < 0, quadratic expression has greatest value at x = -b/2a. This greatest value is given by 4ac - b²/4a = -D/4a.

Answer 2

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