Question

If alpha and beta are the zeros of x square -p(x + 2) - c then alpha + 2 into beta + 2 is equal to?Hint-Ans in terms of c​

Answer 1

Answer:

The answer is 3.

Step-by-step explanation:

$Given \: that \\ f(x) = {x}^{2} - p(x + 2) - c \\ = {x}^{2} - px - 2p - c$

By comparing with, ${ax}^{2} + bx + c$ we have,

$a = 1 \\ b = - p \\ c = (2p + c)$

Sum of zeroes

$( \alpha + \beta ) = \frac{ - b}{a} = \frac{ - ( - p)}{1} = p$

Product of zeroes

$( \alpha \beta ) = \frac{c}{a} = \frac{ - (2p + c)}{1} = (2p + 1)$

Now,

$( \alpha + 2)( \beta + 2) = \alpha \beta + 2( \alpha + \beta ) + 4 \\ = - (2p + 1) + 2(p) + 4 \\ = - 2p - 1 + 2p + 4 \\ = 3$

Ans: 3

Answer 2

$\textsf{\large{\underline{Solution}:}}$

Given That:

$\rm \longrightarrow f(x) = {x}^{2} - p(x + 2) - c$

$\rm \longrightarrow f(x) = {x}^{2} - px - 2p - c$

$\rm \longrightarrow f(x) = {x}^{2} - px - (2p + c)$

Comparing f(x) with Ax² + Bx + C, we get:

$\rm \longrightarrow A= 1$

$\rm \longrightarrow B= -p$

$\rm \longrightarrow C = - 2p-c$

It's given that α and β are the zeros of f(x). Therefore:

$\rm \longrightarrow \alpha + \beta = \dfrac{ - B}{A}$

$\rm \longrightarrow \alpha + \beta = p$

$\rm \longrightarrow \alpha\beta = \dfrac{C}{A}$

$\rm \longrightarrow \alpha\beta = - 2p - c$

Now:

$\rm = ( \alpha + 2)( \beta + 2)$

$\rm = \alpha \beta + 2( \alpha + \beta ) + 4$

$\rm = - 2p - c+ 2p+ 4$

$\rm =4 - c$

★ Which is our required answer.

$\textsf{\large{\underline{Learn More}:}}$

1. Relationship between zeros and coefficients (Quadratic Polynomial)

Let f(x) = ax² + bx + c and let α and β be the zeros of f(x).

Therefore:

$\rm\longrightarrow\alpha+\beta=\dfrac{-b}{a}$

$\rm\longrightarrow\alpha\beta=\dfrac{c}{a}$

2. Relationship between zeros and coefficients (Cubic Polynomial)

Let f(x) = ax³ + bx² + cx + d and let α, β and γ be the zeros of f(x).

Therefore:

$\rm\longrightarrow \alpha+\beta+\gamma=\dfrac{-b}{a}$

$\rm\longrightarrow \alpha\beta+\beta\gamma+\alpha\gamma=\dfrac{c}{a}$

$\rm\longrightarrow \alpha\beta\gamma=\dfrac{-d}{a}$