Question

If $ \alpha,\beta are the roots of the equation x^{2}+x+1 = 0 and \dfrac{\alpha}{\beta} , \dfrac{\beta}{\alpha} are the roots of the equation x^{2}+px+q=0 $ then p equals
A.-1
B.1
C.-2
D.2

Answer 1

Hint: First of all we have to find the sum and product of the given equation {{x}^{2}}+x+1=0 and then by using the given condition of $\dfrac{\alpha }{\beta }$, $\dfrac{\beta }{\alpha }$ which are roots of the equation ${{x}^{2}}+ p x + q=0$ we will find sum and product of the roots of this equation also and then by solving this equation we will get value of p and q



Formula Used:In this question we will be using two formulas which are sum and product of roots of equation
Sum of roots = $\alpha +\beta =-\dfrac{b}{a}$
Product of roots = $\alpha \times\beta =\dfrac{c}{a} $



Complete step by step solution:We are given that,
\alpha,\beta are the roots of the equation ${{x}^ {2}} +x+1$

As the general formula suggests that the general linear equation is $a{{x}^ {2}} +b x+ c $ So here a is 1, b is -1 and c is 1.
therefore according to the formula, sum of roots = $\alpha +\beta =-\dfrac{b}{a} =-\dfrac {(1)}{1} =-1$
And product of roots = $\alpha \times \beta =\dfrac{c}{a} =\dfrac {1}{1} =1$
And as given in the question$ \dfrac {\alpha} {\beta} , \dfrac {\beta} {\alpha} are roots of the given equation {{x}^ {2}} +p x+ q$

therefore product of roots =$ \dfrac{\alpha }{\beta }\times\dfrac{\beta }{\alpha }=\dfrac{c}{a}$ (Here a is 1, b is p and c is q)
therefore$ \dfrac {\alpha} {\beta} \times\dfrac {\beta} {\alpha} =\dfrac{q}{1} $

Now by solving this equation we get q=1 as $\alpha and \beta$ will cancel with each in the L.H.S.
therefore q=1
Now sum of roots = $\dfrac {\alpha} {\beta}+\dfrac {\beta} {\alpha} =-\dfrac{b}{a}$
therefore$ \dfrac {\alpha} {\beta} $+ $\dfrac {\beta} {\alpha} $=-$\dfrac{p}{1} $ (as b is p and a is 1)
therefore $\dfrac {{{\alpha} ^ {2}} + {{\beta} ^ {2}}} {\alpha \times\beta} =-p$
Now as we know the formula for ${{(a+b)} ^ {2}} ={{a}^ {2}} +{{b}^ {2}} +2ab$
If we rearrange it then it comes out to be ${{a}^ {2}} +{{b}^ {2}} ={{(a+b)} ^ {2}}-2ab$
So put this in above equation we get
$\dfrac {{{(\alpha +\beta)} ^ {2}}-2\alpha \beta} {\alpha \times\beta} =-p$
We get the values for $\alpha \times b=1 and \alpha +b=-1$ from the first given equation
So put these two values in the above equation we get
$\dfrac {{{(-1)} ^ {2}}-2(1)}{1} =-p$
$-p=\dfrac {1-2}{1} $
therefore p= 1



Option ‘B’ is correct



Note: Always remember the formulas for sum and product of roots of the equation and in this question we should remember the given condition as from that only we get our value of p and q.