Question

If alpha and beta are the roots of x square - 2 X + 5 equal to zero find the value of Alpha by beta + beta by Alpha

Answer 1

Given:

α and β are the zeroes of the polynomial x² - 2x + 5.

• a = 1

• b = -2

• c = 5

To find out:

Find the value of α/β + β/α ?

Solution:

★Sum of zeroes:

α + β = -b/a

⇒ - ( - 2) / 1

⇒ 2

★Product of zeroes:

αβ = c/a

⇒ 5/1

⇒5

Now,

α/β + β/α [ Given ]

⇒ α² + β² /αβ

We know that,

α² + β² = ( α + β )² - 2αβ

⇒ ( α + β )² - 2αβ / αβ

☆Putting the values, we get

⇒ ( 2 )² - 2 × 5 / 5

⇒ 4 - 10/5

⇒ -6 / 5

Thus ,the value of α/β + β/α is -6/5 .

Answer 2

GIVEN :–

• A quadratic equation have two roots $\: \: \sf \alpha \: \: and \: \beta \: \:$

TO FIND :–

$\\ \: \: \sf \dfrac{ \alpha }{ \beta } + \dfrac{ \beta }{ \alpha } = ?$

SOLUTION :–

$\\ \sf \implies {x}^{2} - 2x + 5 =0$

$\\ \sf \longrightarrow \: \: sum \: \: of \: \: roots = - \dfrac{coffieciant \: \: of \: \: x}{coffieciant \: \: of \: \: {x}^{2} }$

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

$\\ \sf \implies \: \: \alpha + \beta = 2 \: \: \: - - - eq.(1)$

$\\ \sf \longrightarrow \: \: product \: \: of \: \: roots = \dfrac{constant \: \: term}{coffieciant \: \: of \: \: {x}^{2} }$

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

$\\ \sf \implies \: \: \alpha \beta = 5 \: \: \: - - - eq.(2)$

• Now Let's find –

$\\ \: \: \sf = \dfrac{ \alpha }{ \beta } + \dfrac{ \beta }{ \alpha }$

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

$\\ \: \: \sf = \dfrac{ {( \alpha + \beta )}^{2} - 2 \alpha \beta }{ \alpha \beta } \: \: \: \: \: \: \:[ \: \because \: {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab ]$

• Now using eq.(1) and eq.(2) –

$\\ \: \: \sf = \dfrac{ {(2 )}^{2} - 2 (5) }{5 }$

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

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

• Hence , $\: \: \sf \dfrac{ \alpha }{ \beta } + \dfrac{ \beta }{ \alpha } = - \dfrac{6}{5}$