Question

If alpha and beta are the roots of the equation x^2 -3x +1 = 0, then find the equation with roots 1/ alpha -2,1/beta -2
please give full solution ​

Answer 1

$\huge\mathfrak\blue{Answer:}$

Given:

• We have been given a Quadratic Polynomial x² - 3x + 1 = 0
• α and β are the zeros of given Quadratic Polynomial

To Find:

• To find a Quadratic Polynomial whose zeros are (1/α-2) and (1/β-2)

Solution:

We have been given a Quadratic Polynomial

$\boxed{\text{f ( x ) = x² - 3x + 1}}$

Comparing it with standard form of Equation

$\boxed{\text{a = 1 and b = ( - 3 ) and c = 1}}$

______________________________

Relation Between Roots and Coefficient of a Quadratic Polynomial

$\bigstar \: \text{α + β =}$$\sf{- \dfrac{b}{a}}$

$\implies \text{α + β = }$ $\sf{- \left ( \dfrac{-3}{1} \right ) }$

$\implies \boxed{\text{α + β = 3}}$ ------------------ 1

Also we know that

$\bigstar \: \text{αβ =}$$\sf{\dfrac{c}{a}}$

$\implies \text{αβ =}$$\sf{\dfrac{1}{1}}$

$\implies \boxed{\text{αβ = 1 }}$ ---------------------- 2

_________________________________

We have to find a Quadratic Polynomial whose zeros are (1 / α - 2) and ( 1 / β - 2)

Finding sum and product of Required zeros

$\odot$ Sum of Zeros

$\implies \dfrac{1}{\text{α - 2}} + \dfrac{1}{\text{β - 2}}$

Taking LCM

$\implies \dfrac{1}{\text{α - 2}} + \dfrac{1}{\text{β - 2}}$

$\implies \dfrac{\text{β - 2 + α - 2}}{\text{(α - 2)(β - 2)} }$

$\implies \dfrac{\text{α + β - 4}}{\text{αβ - 2α - 2β + 4} }$

$\implies \dfrac{\text{( α + β ) - 4}}{\text{αβ - 2 ( α + β ) + 4}}$

Substituting the Values in above Equation

$\implies \dfrac{\text{( 3 ) - 4}}{\text{1 - 2 ( 3 ) + 4}}$

$\implies \dfrac{\text{-1}}{\text{-1}}$

$\implies \text{1 }$

$\large\boxed{\text{Sum of Zeros = 1 }}$

___________________________

$\odot$ Product of Zeros

$\implies \dfrac{1}{\text{α - 2}} \times \dfrac{1}{\text{β - 2} }$

$\implies \dfrac{1}{\text{(α - 2)(β - 2)}}$

$\implies \dfrac{1}{\text{αβ - 2α - 2β + 4}}$

$\implies \dfrac{1}{\text{αβ - 2 ( α + β ) + 4}}$

Substituting the Value in above Equation

$\implies \dfrac{1}{\text{1 - 2 ( 3 ) + 4}}$

$\implies \dfrac{1}{-1}$

$\large\boxed{\text{Product of Zeros = ( - 1 ) }}$

________________________________

Quadratic Equation whose sum and product of zeros is given

$\text{p ( x ) = k [ x² - (Sum) x + Product]}$

$\text{p ( x ) = k [ x² - ( 1 ) x + ( -1 ) ]}$

$\text{p ( x ) = k [ x² - x - 1]}$

Where k is a non zero Arbitrary Constant

______________________________

$\huge\underline{\sf{\red{A}\orange{n}\green{s}\pink{w}\blue{e}\purple{r}}}$

$\large\boxed{\text{ p ( x ) = k [ x² - x - 1 ]}}$

_______________________________