Question

If α and β are the roots of the equation x² + 5x + 5 =0 . Then Find the Equation whose roots are (α + 1) and (β + 1).​

Answer 1

Answer:

$\LARGE\underbrace{\purple{\sf{Required \; Solution:}}}$

The quadratic equation whose roots are α + 1, β + 1 is x² + 3x + 1 = 0

Given equation is

 ⠀⠀⠀⠀$\sf{x^2 + 5x + 5 = 0}$ ——— (1)

Comparing with ax² + bx + c = 0, we get

 ⠀⠀⠀⠀$\sf{a = 1, \; b = 5, \; c = 5}$

Roots of (1) are α, β

 ⠀⠀⠀⠀$\sf{sum \; of \; roots = \alpha + \beta = \dfrac{-b}{a} = \dfrac{-5}{1} = -5}$ ——— (2)

 ⠀⠀⠀⠀$\sf{product \; of \; roots = \alpha\beta = \dfrac{c}{a}=\dfrac{5}{1} = 5}$——— (3)

Quadratic equation with roots α + 1, β + 1 is given by

 ⠀⠀⠀⠀x² - (sum of roots)x + (product of roots) = 0

 ⠀⠀⠀⠀x² - (α + 1 + β + 1)x + (α+1)(β+1) = 0

 ⠀⠀⠀⠀x² - (α + β + 2)x + αβ + α + β + 1 = 0

 ⠀⠀⠀⠀x² - (-5+2)x + 5 + (-5) + 1 = 0       [from (2) and (3)]

 ⠀⠀⠀⠀x² - (-3)x + 5 - 5 + 1 = 0

 ⠀⠀⠀⠀x² + 3x + 1 = 0

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Answer 2

Answer:

$\large \bf \clubs \:  Given  :-$

α and β are the roots of the equation

x² + 5x + 5 =0

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

The Equation whose roots are  (α + 1)  and (β + 1).

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

1》For a qudratic Equation of the Form ax² + bx + c = 0

Sum of Roots = $\sf-\dfrac{b}{a}$

Product of Roots = $\sf\dfrac{c}{a}$

2》 A Quadratic Equation whose sum and product of Roots are S and P respectively is given by x² - Sx + P = 0

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$\large \bf \clubs \:  Solution :-$

As α and β are the roots of the equation

x² + 5x + 5 =0

Hence ,

$\pmb{ \alpha  +  \beta  = - 5 }  \:  \:  -  -  -  - (1)\\  \bf and \\  \pmb{ \alpha  \beta  = 5} \:  \:  -  -  -  - (2)$

For The  Equation whose Roots are

(α + 1)  and (β + 1).

$\sf S um \:  of \:  Roots  = S  \\  \\   = \sf \alpha  + 1 +  \beta  + 1 \\  \\  =  \alpha  +  \beta  + 2 \\  \\  \bf \: \:  \:  \{using \:  \: (1)  \} \\  \\  \sf S =  - 5 + 2\\  \\ \large:\longmapsto \boxed{  \pmb { \boxed{S =  - 3}}}$

$\sf P roduct  \: of \:  Roots  = P \\  \\  = ( \alpha  + 1)( \beta  + 1) \\  \\  =  \alpha  \beta  +  \alpha  +  \beta  + 1 \\  \\  \bf \:  \:  \:  \{using  \: (1) \: and \: (2) \} \\  \\  \sf P = 5 - 5 + 1 \\  \\ \large :\longmapsto\boxed{ \pmb{ \boxed{P = 1}}}$

Hence The  Equation whose Roots are

(α + 1)  and (β + 1) will be x² - Sx + P = 0

Where S and P are Sum and Product of roots.

That is ,

The Equation is x² - ( - 3) x + 1 =0

$\large \pink{ : \longmapsto \bf  {x}^{2} + 3x + 1 = 0 }$

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

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