Question

If alpha and beta are roots of x²-5x+6=0.construct a quadratic equation whose roots are 1÷alpha and 1÷beta

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\: \alpha , \beta \: are \: roots \: of \: {x}^{2} - 5x + 6 = 0$

We know that,

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

$\bf\implies \: \alpha + \beta = - \dfrac{( - 5)}{1} = 5$

and

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$

$\bf\implies \: \alpha \beta = \dfrac{6}{1} = 6$

Now,

We have to form a quadratic equation whose roots are

$\: \sf \: \dfrac{1}{ \alpha } ,\dfrac{1}{ \beta }$

Consider,

Sum of roots, S

$\rm \: = \: \: \dfrac{1}{ \alpha } + \dfrac{1}{ \beta }$

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

$\rm \: = \: \dfrac{5}{6}$

$\rm :\implies\: \boxed{ \bf{S = \frac{5}{6}}}$

Now,

Product of roots, P

$\rm \: = \: \: \dfrac{1}{ \alpha } \times \dfrac{1}{ \beta }$

$\rm \: = \: \dfrac{1}{ \alpha \beta }$

$\rm \: = \: \dfrac{1}{6}$

$\rm :\implies\: \boxed{ \bf{P = \frac{1}{6}}}$

Now,

Required quadratic equation is given by

$\rm :\longmapsto\: {x}^{2} - Sx + P = 0$

$\rm :\longmapsto\: {x}^{2} - \dfrac{5}{6} x + \dfrac{1}{6} = 0$

$\bf\implies \: {6x}^{2} - 5x + 1 = 0$

Remark :-

Short cut trick :-

$\rm :\longmapsto\: \alpha , \beta \: are \: roots \: of \: {ax}^{2} + bx + c= 0 \: then \:$

$\rm \: quadratic \: equation \: whose \: roots \: are \: \dfrac{1}{ \alpha },\dfrac{1}{ \beta } \: is \: given \: by\:$

$\rm \: {cx}^{2} + bx + a = 0$

Answer 2

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▪ Given :-

$\large\sf{\alpha \: and \: \beta \: are \: the \: zeros \: of \: eq {}^{n} } : \\ \bf{x}^{2} - 5x + 6 = 0$

▪ To Find :-

$\large\sf{A \: Quadratic \: Eq {}^{n} \: whose \: zeros \: are}: \\ \frac{1}{ \alpha } \: \bf and \: \frac{1}{ \beta }$

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▪ Concept To Mind :-

$\large \pmb{\mathfrak{For \: \: \text{A} \: \: Qudratic \: \: Eq^n \: \: of \: \: the \: \: Form}} :$

$\Large\bf a {x}^{2} + bx + c$

$\large\sf{Sum \: of \: Zeros = - \dfrac{b}{a} } \\ \\ \large \sf{ Product \: of \: Zeros = \frac{c}{a} }$

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▪ Solution :-

$\large \sf As \: \mathtt{\alpha \: and \: \beta \: are \: the \: zeros \: of \: eq {}^{n} } : \\ \bf{x}^{2} - 5x + 6 = 0 \\ \\ \Large : \longmapsto \: \alpha + \beta = 5 \: \: \: \: - - - - (1)$

$\huge \pmb{ \mathfrak{ \text{A}lso,}}$

$\Large\alpha \beta = 6 \: \: \: \: - - - - (2)$

For The Quadratic Equation whose Zeros are $\dfrac{1}{ \alpha } \: \bold{and} \: \dfrac{1}{ \beta }$

$\sf Sum \: of \: Zeros = S = \dfrac{1}{ \alpha } + \dfrac{1}{ \beta } \\ \\ \large \sf = \frac{ \beta + \alpha }{ \alpha \beta } \\ \\ \Large\pmb{ \boxed{ \boxed{ S = \frac{5}{6} }}}$

$\sf Product \: of \: Zeros = P = \dfrac{1}{ \alpha } \times \dfrac{1}{ \beta } \\ \\ = \large\frac{1}{ \alpha \beta } \\ \\ \Large\pmb{ \boxed{ \boxed{P = \frac{1}{6} }}}$

Now, Required Equation will be :

$\Large\mathtt{ {x}^{2} - S \: x + P = 0 }$

That is :-

$\mathtt{{x}^{2} - \dfrac{5}{6} x + \frac{1}{6} = 0} \\ \\ : {\longmapsto \pmb{6 {x}^{2} - 5x + 1 = 0}}$

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

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