Question

if alfa and beta are the zeroes of quadratic polonomial x2-5x+4 then the value of alfa/beta and beta/alfa is

Answer 1

Correct Question:

If $\sf{\alpha}$ and $\sf{\beta}$ are the zeroes of quadratic polynomial x² - 5x + 4, then find the value of $\sf{\dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha}}$.

Step-by-step explanation:

Given :-

• $\sf{\alpha}$ and $\sf{\beta}$ are the zeroes of quadratic polynomial x² - 5x + 4.

To find :-

• Value of $\sf{\dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha}}$.

Solution :-

Quadratic polynomial :

x² - 5x + 4 = 0..............(i)

On comparing eq(i) with ax² + bx + c = 0, we get,

• a = 1
• b = -5
• c = 4

We know,

• ${\boxed{\sf{Sum\:of\: zeroes=\dfrac{-b}{a}}}}$

• ${\boxed{\sf{Product\:of\: zeroes=\dfrac{c}{a}}}}$

Then,

$\sf \: \alpha + \beta = \dfrac{ - ( - 5)}{1} = 5$

&

$\sf \: \alpha \beta = \dfrac{4}{1} = 4$

Now find the value of $\sf{\dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha}}$.

$\sf \: = \dfrac{ \alpha }{ \beta } + \dfrac{ \beta }{ \alpha } \\ \\ = \sf \dfrac{ { \alpha }^{2} + { \beta }^{2} }{ \alpha \beta } \\ \\ = \sf \dfrac{ {( \alpha + \beta )}^{2} - 2 \alpha \beta }{ \alpha \beta } \\ \\ = \sf \dfrac{ {5}^{2} - 2 \times 4}{4} \\ \\ \sf \: = \dfrac{25 - 8}{4} \\ \\ \sf \: = \dfrac{17}{4}$

Therefore, the value of $\sf{\dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha}}$ is 17/4.