Question

If alpha and beta are the zeroes of the quadratic polynomial 5x^2 - 16x - 45, then find the value of alpha/beta + beta/alpha

Answer 1

Given

Zeros of polynomial are a & ß

Quadratic polynomial = 5x² - 16x - 45

To Find

Value of : a/ß + ß/a

Solution

Here, the zeros are a & ß.

We can find value of a & ß by factorization method :

➝ 5x² - 16x - 45 = 0

➝ 5x² - 25x + 9x - 45 = 0

➝ 5x(x - 5) + 9(x - 5) = 0

➝ (5x + 9)(x - 5) = 0

➝ 5x + 9 = 0 or, x - 5 = 0

➝ 5x = -9 or, x = 5

➝ x = -9/5 or, x = 5

Therefore,

Zeros of polynomial are : -9/5 & 5

Here,

• a = -9/5
• ß = 5

We have to find value of (a/ß) + (ß/a)

Putting values we get :

⇒ a/ß + ß/a = (-9/5)/5 + 5/(-9/5)

⇒ a/ß + ß/a = -9/25 + (25/-9)

⇒ a/ß + ß/a = (81 + 625)/-225

⇒ a/ß + ß/a = (706)/-225

⇒ a/ß + ß/a = -706/225

Therefore,

Required value = -706/225

Answer 2

Answer:

$Given \: \alpha \: and \: \beta \: are \: the \: zeros \: of \: quadratic \: poynomial.$

Given polynomial f(x) =5x²-16x-45.

On solving this quadratic equation by middle term factorization method ;

5x²-16x-45 =0

5x²-25x+9x-45=0

5x(x-5)+9(x-5)=0

(5x+9)(x-5)=0

x = - 9/5 or 5. [zeros of the polynomial]

$And \: so, \alpha = - \frac{9}{5} \: and \: \beta = 5.$

To Find :

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

$= > \frac{ - \frac{9}{5} }{5} + \frac{5}{ - \frac{9}{5} }$

$= > ( - \frac{9}{5} \times \frac{1}{5} ) + ( 5 \times - \frac{ 5}{9} )$

$= >( - \frac{9}{25} - \frac{25}{9})$

$= >( - \frac{706}{225} )$

$Therefore, \: the \: value \: of \: \frac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha } \: is \: ( - \frac{706}{225} ).$