Question

If a, B are the roots of the equation x2 -2x + 3 = 0, form the equation whose roots are a + 1 /beta, beta + 1/alpha​

Answer 1

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

Given that,

$\rm :\longmapsto\: \alpha , \beta \: are \: the \: roots \: of \: equation \: {x}^{2} - 2x + 3 = 0$

We know,

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

$\rm\implies \: \alpha + \beta = - \dfrac{( - 2)}{1} = 2$

And,

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

$\rm\implies \: \alpha \beta = \dfrac{3}{1} = 3$

Now, Consider

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

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

$\rm \:  =  \: 2 + \dfrac{ \beta + \alpha }{ \alpha \beta }$

$\rm \:  =  \: 2 + \dfrac{2}{3}$

$\rm \:  =  \: \dfrac{6 + 2}{3}$

$\rm \:  =  \: \dfrac{8}{3}$

Now, Consider

$\rm :\longmapsto\:P = \bigg[ \alpha + \dfrac{1}{ \beta } \bigg] \times \bigg[ \beta + \dfrac{1}{ \alpha } \bigg]$

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

$\rm \:  =  \: 3 + 2 + \dfrac{1}{3}$

$\rm \:  =  \: 5 + \dfrac{1}{3}$

$\rm \:  =  \: \dfrac{15 + 1}{3}$

$\rm \:  =  \: \dfrac{16}{3}$

So, Required Quadratic equation is given by

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

$\rm :\longmapsto\: {x}^{2} - \dfrac{8}{3}x + \dfrac{16}{3} = 0$

$\rm :\longmapsto\: \dfrac{ {x}^{2} - 8x + 16}{3} = 0$

$\rm :\longmapsto\: {x}^{2} - 8x + 16 = 0$

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Additional Information :-

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

If Discriminant, D > 0, then roots of the equation are real and unequal.

If Discriminant, D = 0, then roots of the equation are real and equal.

If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

Discriminant, D = b² - 4ac