Question

If \[\alpha ,\beta \] are the roots of the equation \[{x^2} - px + q = 0\], then the values of \[(\alpha + \beta )x - ({\alpha ^2} + {\beta ^2}){x^2}/2 + ({\alpha ^3} + {\beta ^3}){x^3}/3 + ...\] is
A. \[\log (1 - px + q{x^2})\]
B. \[\log (1 + px - q{x^2})\]
C. \[\log (1 + px + q{x^2})\]
D. None of these

Answer 1

Hint: In the given question the roots of a quadratic equation is given and we are asked to find the value of \[(\alpha + \beta )x - ({\alpha ^2} + {\beta ^2}){x^2}/2 + ({\alpha ^3} + {\beta ^3}){x^3}/3 + ...\]. Quadratic equation is an equation with one variable with \[2\] as the maximum power of the variable. Standard form of an quadratic equation is: \[a{x^2} + bx + c = 0\], where \[a,b\] and \[c\] are the coefficients and the quadratic equation is roots formula is: \[{x^2} - ({\text{sum of the roots}}) + ({\text{product of the roots}}) = 0\], by using this formula let solve our question.

Complete step by step answer:
Given quadratic equation is: \[{x^2} - px + q = 0\]
Roots of the quadratic equation is: \[\alpha ,\beta \]
We have the quadratic equation is roots formula is: \[{x^2} - ({\text{sum of the roots}}) + ({\text{product of the roots}}) = 0\], by applying this in \[{x^2} - px + q = 0\],
Sum of the roots: \[\alpha + \beta = p\]
Product of the roots: \[\alpha \beta = q\]
\[(\alpha + \beta )x - ({\alpha ^2} + {\beta ^2}){x^2}/2 + ({\alpha ^3} + {\beta ^3}){x^3}/3 + ...\]
From the second term here is a fraction with it let rewrite with representing fraction in front,
\[(\alpha + \beta )x - \dfrac{1}{2}({\alpha ^2} + {\beta ^2}){x^2} + \dfrac{1}{3}({\alpha ^3} + {\beta ^3}){x^3} + ...\]

Now multiplying the values inside brackets we will get,
\[\left[ {\alpha x + \beta x} \right] - \left[ {\dfrac{1}{2}\alpha {x^2} + \dfrac{1}{2}\beta {x^2}} \right] + \left[ {\dfrac{1}{3}\alpha {x^3} + \dfrac{1}{3}\beta {x^3}} \right] + ...\]
Remove the brackets and also multiply the \[ - \] inside the bracket of second term,
\[ \alpha x + \beta x - \dfrac{1}{2}\alpha {x^2} - \dfrac{1}{2}\beta {x^2} + \dfrac{1}{3}\alpha {x^3} + \dfrac{1}{3}\beta {x^3} + ...\]
By separately writing the \[\alpha \] and \[\beta \] values,
\[ \left[ {\alpha x - \dfrac{1}{2}\alpha {x^2} + \dfrac{1}{3}\alpha {x^3} + ...} \right] + \left[ {\beta x - \dfrac{1}{2}\beta {x^2} + \dfrac{1}{3}\beta {x^3} + ...} \right]\]
We can clearly see that it is in the form \[x - \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{3} + ...\] which can be shortly written with the help of logarithms.
\[\log (1 - \alpha x) + \log (1 - \beta x)\]

Now by applying product rule of logarithm, \[\log (ab) = \log (a) + \log (b)\]
\[\log \left( {(1 - \alpha x)(1 - \beta x)} \right)\]
Multiplying the brackets inside log,
\[\log (1 - \alpha x - \beta x + \alpha \beta {x^2})\]
Taking \[ - x\] common in \[ - \alpha x\] and \[ - \beta x\],
\[\log (1 - x(\alpha + \beta ) + \alpha \beta {x^2})\]
\[\Rightarrow \log (1 - (\alpha + \beta )x + \alpha \beta {x^2})\]
Already we have found that sum of the roots, \[\alpha + \beta = p\] and product of the roots \[\alpha \beta = q\]by substituting this,
\[ \therefore \log (1 - px + q{x^2})\]

Hence option A is correct.

Note:Logarithm is a quantity representing the power to which a fixed number (the base) must be raised to produce a given number. Expansion of \[\log (1 + x) = x - \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{3} - \dfrac{{{x^4}}}{4} + ........... + \dfrac{{{x^n}}}{n}\]. Product rule of logarithms says that the logarithm of a product of two quantities is the sum of the logarithms of the two factors \[\log (ab) = \log (a) + \log (b)\].