Question

IF ALPHA AND BEETA ARE THE ZEROES OF THE QUADRATIC POLYNOMIAL F (X)=X^2-X-4 FIND THE VALUE OF 1/ALPHA+1/BEETA-2ALPHA BEETA​

Answer 1

Answer:

$\displaystyle{\boxed{\red{\sf\:\dfrac{1}{\alpha}\:+\:\dfrac{1}{\beta}\:-\:2\:\alpha\:\beta\:=\:\dfrac{31}{4}}}}$

Step-by-step-explanation:

The given quadratic polynomial is x² - x - 4.

Let the zeroes of the quadratic polynomial be $\displaystyle{\sf\:\alpha\:\&\:\beta}$

Now,

x² - x - 4

Compairing with ax² + bx + c, we get,

• a = 1
• b = - 1
• c = - 4

We know that,

$\displaystyle{\pink{\sf\:Sum\:of\:zeroes\:=\:-\:\dfrac{b}{a}}}$

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

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

$\displaystyle{\implies\boxed{\pink{\sf\:\alpha\:+\:\beta\:=\:1}}}$

Now,

$\displaystyle{\blue{\sf\:Product\:of\:zeroes\:=\:\dfrac{c}{a}}}$

$\displaystyle{\implies\sf\:\alpha\:.\:\beta\:=\:\dfrac{-\:4}{1}}$

$\displaystyle{\implies\boxed{\blue{\sf\:\alpha\:.\:\beta\:=\:-\:4}}}$

We have to find the value of,

$\displaystyle{\sf\:\dfrac{1}{\alpha}\:+\:\dfrac{1}{\beta}\:-\:2\:\alpha\:\beta}$

$\displaystyle{\implies\sf\:\dfrac{\alpha\:+\:\beta}{\alpha\:.\:\beta}\:-\:2\:\times\:\alpha\:.\:\beta}$

$\displaystyle{\implies\sf\:\dfrac{1}{-\:4}\:-\:2\:\times\:(\:-\:4\:)}$

$\displaystyle{\implies\sf\:-\:\dfrac{1}{4}\:+\:8}$

$\displaystyle{\implies\sf\:-\:\dfrac{1\:+\:8\:\times\:4}{4}}$

$\displaystyle{\implies\sf\:-\:\dfrac{1\:+\:32}{4}}$

$\displaystyle{\implies\sf\:\dfrac{31}{4}}$

$\displaystyle{\therefore\:\underline{\boxed{\red{\sf\:\dfrac{1}{\alpha}\:+\:\dfrac{1}{\beta}\:-\:2\:\alpha\:\beta\:=\:\dfrac{31}{4}}}}}$