Question

the quadratic equation whose roots are p and 1/p is given by​

Answer 1

$\large \bf \clubs \:  Given :-$

For a Quadratic Equation :

   

• First Root = p

• Second Root = 1/p

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$\large \bf \clubs \:   To  \: Find :-$

• The Quadratic Equation .

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$\large \bf \clubs \:   Main  \:  Concept : -$

✏ If sum and product of zeros of any quadratic Equation are A and B respectively,

Then,

The quadratic Equation is given by :-

$\bf  {x}^{2}  - A \: x +B=0$

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$\large \bf \clubs \:  Solution  :-$

Here,

$\sf Sum = A = p + \frac{1}{p} \\ \\ :\longmapsto \bf Sum = A = \frac{ {p}^{2} + 1}{p} \\ \bf and\\ \sf Product = B= p \times \frac{1}{p} \\ \\ \bf :\longmapsto Product = B = 1$

So,

Required Equation should be :

$\bf  {x}^{2}  - A \: x + B=0$

$:\longmapsto  \tt{x}^{2}  - \bigg(\dfrac{p^2 +1}{p} \bigg)x + 1=0$.

$\large\purple{:\longmapsto\pmb{p {x}^{2}  -(p^2+1)x +p=0}}$

$\Large \red{\mathfrak{  \text{W}hich \:   \: is  \:  \: the  \:  \: required} }\\ \huge \red{\mathfrak{ \text{ A}nswer.}}$

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Answer 2

$\huge { \red{ \green{ \blue{ \pink{ \orange{\boxed{ \mathbb \blue{p {x}^{2} - ( {p}^{2} + 1)x + p}}}}}}}}$

Step-by-step explanation:

QUESTION :-

The quadratic equation whose roots are p and 1/p is given by -

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SOLUTION :-

Let the two roots as α & β respectively.

So,

α = p

β = 1/p

Now,

Sum of roots (α + β) = $\bf p + \dfrac{1}{p}$

$= > \bf \dfrac{ {p}^{2} + 1}{p}$

=> Sum of roots (S) = $\bf \dfrac{{p}^{2} + 1}{p}$

And,

Product of Roots (αβ) = $\bf \cancel{p} \times \dfrac{1}{\cancel{p}}$

=> 1

=> Product of roots (P) = 1

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To find the quadratic equation, we use formula,

x² -(S)x + P ......[S = α+β, P = αβ]

[put the value of S and P]

$= > \bf{x}^{2} - ( \frac{ {p}^{2} + 1 }{p} )x + 1$

[multiply the equation by p]

$= > \bf p( {x}^{2} - ( \frac{ {p}^{2} + 1}{p} )x + 1)$

=> px² - (p² + 1)x + p

So,

the required quadratic equation is px² - (p² + 1)x + p.

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Hope it helps.

#BeBrainly :-)