Question

if one zero of p(x) 5x²+13x+k is reciprocal of other then value of k is​

Answer 1

Answer:

alpha×(1/alpha)=k/5

1=k/5

k=5

Answer 2

Given :-

One zero of p ( x ) = 5x² + 13x + k is reciprocal of other

To Find :-

Value of k

Solution :-

Before starting the question , let's recall Vieta Formula for a ${\pmb { \blue { \bf { Quadratic \: Polynomial }}}}$

If f ( x ) = ax² + bx + c is the quadratic polynomial with zeros/roots $\alpha$ and $\beta$ . Then ,

• $\sf \alpha + \beta = - \dfrac{b}{a}$

• $\sf \alpha \cdot \beta = \dfrac{c}{a}$

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Now , comparing p ( x ) with the standard form of a quadratic polynomial , we have ;

• a = 5
• b = 13
• c = k

Also , let us assume that , $\alpha$ and $\beta$ are zeroes of p ( x ) . Since , it is given that , 1 zero is reciprocal of other . Which implies ;

• $\beta = \dfrac{1}{\alpha}$

Also ;

$\quad \leadsto \quad \sf \alpha \times \beta = \dfrac{c}{a}$

Putting the values , we have ;

${ : \implies \quad \sf \alpha \times \dfrac{1}{\alpha} = \dfrac{k}{5}}$

${ : \implies \quad \sf \cancel{\alpha} \times \dfrac{1}{\cancel{\alpha}} = \dfrac{k}{5}}$

${ : \implies \quad \sf \dfrac{k}{5} = 1 }$

${ : \implies \quad \bf \therefore \quad k = 5 }$

Henceforth , The Required Answer is 5 :)