Question

Answer the following question
1)If alpha & beta are the roots of x square -2x +5 equal to zero find the value of alpha by beta + beta by alpha?
solve this fast ヘ(^_^)ヘ​

Answer 1

$\underline\mathfrak{Given:-}$

• A quadratic equations have two roots alpha and beta.

$\underline\mathfrak{To \: \: Find:-}$

• Find the alpha/beta + beta/alpha ...?

$\underline\mathfrak{Solutions:-}$

• The equation is in the form of:-

x² - 2x + 5 = 0

• a = 1
• b = -2
• c = 5

$\: \: \: \: \: \therefore {Sum \: \: of \: \: zeroes} \: \: = \: \: \frac{ \: - \: coefficient \: \: of \: \: x}{coefficient \: \: of \: \: {x}^{2}}$

$\: \: \: \: \: \leadsto \: \: \alpha \: \: + \: \beta \: \: = \: \: - \: \frac{(-2)}{1}$

$\: \: \: \: \: \leadsto \: \: \alpha \: \: + \: \beta \: \: = \: \: {2} \: \: \: \: \: ....{(i)}.$

$\: \: \: \: \: \therefore {Product \: \: of \: \: zeroes} \: \: = \: \: \frac{constant \: \: term}{coefficient \: \: of \: \: {x}^{2}}$

$\: \: \: \: \: \leadsto \: \: \alpha\beta \: \: = \: \: \frac{5}{1}$

$\: \: \: \: \: \leadsto \: \: \alpha\beta \: \: = \: \: {5} \: \: \: \: .....{(ii)}.$

Now,

$\: \: \: \: \: \leadsto \: \: \frac{\alpha}{\beta} \: + \: \frac{\beta}{\alpha}$

$\: \: \: \: \: \leadsto \: \: \frac{{\alpha}^{2} \: + \: {\beta}^{2}}{\alpha\beta}$

$\: \: \: \: \: \leadsto \: \: \frac{{({\alpha} \: + \: {\beta})}^{2} \: - \: {{2} \: \alpha\beta}}{\alpha\beta} \: \: \: \: \: {[{(a + b)}^{2} \: \: = \: \: {a}^{2} \: + \: {b}^{2} \: + \: {2} \: \alpha\beta]}$

Now, using Eq. (i) and Eq. (ii).

$\: \: \: \: \: \leadsto \: \: \frac{{(2)}^{2} \: - \: {2} \: \times \: {5}}{5}$

$\: \: \: \: \: \leadsto \: \: \frac{{4} \: - \: {10}}{5}$

$\: \: \: \: \: \leadsto \: \: \frac{-6}{5}$

Hence,

$\: \: \: \: \: \leadsto \: \: \frac{\alpha}{\beta} \: + \: \frac{\beta}{\alpha} \: \: = \: \: \frac{-6}{5}$

Answer 2

$\large\mathcal{\green{\underbrace{\orange{SOLUTION:-}}}}$

GIVEN :-

✍️ $\rm{\red{\alpha}\:and\:\red{\beta}\:}$ are the roots of ‘x² - 2x + 5 = 0’ .

TO FIND :-

• The value of ‘$\rm{\red{\dfrac{\alpha}{\beta}\:+\:\dfrac{\beta}{\alpha}\:}}$’ .

CALCULATION :-

$\checkmark\rm{\purple{Sum\:of\:zeros\:=\:\dfrac{-\:coefficient\:of\:x}{coefficient\:of\:x^2}\:}}$

✍️ Here,

• coefficient of x = -2

• coefficient of x² = 1

• constant term = 5

$\rm{\implies\:\alpha\:+\:\beta\:=\:\dfrac{-(-2)}{1}\:}$

$\rm{\red{\implies\:\alpha\:+\:\beta\:=\:2\:}}$

$\checkmark\:\rm{\purple{Product\:of\:zeros\:=\:\dfrac{constant\:term}{coefficient\:of\:x^2}\:}}$

$\rm{\implies\:\alpha\:\beta\:=\:\dfrac{5}{1}\:}$

$\rm{\red{\implies\:\alpha\:\beta\:=\:5\:}}$

✍️ Now, the value of ‘ $\rm{\dfrac{\alpha}{\beta}\:+\:\dfrac{\beta}{\alpha}\:}$ ’ .

$\rm{=\:\dfrac{\alpha^{2}\:+\:\beta^{2}}{\alpha\:\beta}\:}$

$\rm{=\:\dfrac{(\alpha\:+\:\beta)^2\:-\:2\alpha\beta}{\alpha\beta}\:}$

$\rm{=\:\dfrac{2^2\:-\:2\times{5}}{5}\:}$

$\rm{=\:\dfrac{4\:-\:10}{5}\:}$

$\rm{=\:\dfrac{-6}{5}\:}$

$\bigstar\:\rm{\blue{\boxed{\dfrac{\alpha}{\beta}\:+\:\dfrac{\beta}{\alpha}\:=\:\dfrac{-6}{5}\:}}}$