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Question: If we are given the roots as \(\alpha ,\beta \) of the equation \(a{{x}^{2}}+bx+c=0\) then the roots...

If we are given the roots as α,β\alpha ,\beta of the equation ax2+bx+c=0a{{x}^{2}}+bx+c=0 then the roots of the equation
(a+b+c)x2−(b+2c)x+c=0\left( a+b+c \right){{x}^{2}}-\left( b+2c \right)x+c=0 are
a) α+1α,β+1β\dfrac{\alpha +1}{\alpha },\dfrac{\beta +1}{\beta }
b) α−1α,β−1β\dfrac{\alpha -1}{\alpha },\dfrac{\beta -1}{\beta }
c) αα+1,ββ+1\dfrac{\alpha }{\alpha +1},\dfrac{\beta }{\beta +1}
d) αα−1,ββ−1\dfrac{\alpha }{\alpha -1},\dfrac{\beta }{\beta -1}

Explanation

Solution

We have the roots of the equation and we know that the sum of the roots of an equation =(- Coefficient of x)(Coefficient of    x2)\dfrac{(\text{- Coefficient of x})}{(\text{Coefficient of}\;\;{x^2})} and the product of the roots =(Constant)(Coefficient of    x2)\dfrac{(\text{Constant})}{(\text{Coefficient of}\;\;{x^2})} if the equation is of the form ax2+bx+c=0a{{x}^{2}}+bx+c=0.We will calculate the sum of the roots and product of the roots. We will now take the second equation whose roots we have to calculate. We will use the above formula to get the sum and product. We will now arrange the expression such that we input the values of (b/a) and (c/a) so that we can get the answer in the form of α,β\alpha,\beta .

Complete step-by-step solution:
We have equation, ax2+bx+c=0a{{x}^{2}}+bx+c=0,
We know that the sum of roots = −ba\dfrac{-b}{a}
⇒α+β=−ba ⇒−(α+β)=ba \begin{aligned} & \Rightarrow \alpha +\beta =\dfrac{-b}{a} \\\ & \Rightarrow -\left( \alpha +\beta \right)=\dfrac{b}{a} \\\ \end{aligned}
We know that the product of roots = ca\dfrac{c}{a}
⇒αβ=ca\Rightarrow \alpha \beta =\dfrac{c}{a}
We have equation, (a+b+c)x2−(b+2c)x+c=0\left( a+b+c \right){{x}^{2}}-\left( b+2c \right)x+c=0,
Let the roots of above equation be p and q,
We know that the sum of roots = −ba\dfrac{-b}{a}
⇒p+q=b+2ca+b+c\Rightarrow p+q=\dfrac{b+2c}{a+b+c}
We know that the product of roots = ca\dfrac{c}{a}
⇒pq=ca+b+c\Rightarrow pq=\dfrac{c}{a+b+c}
We have, p+q=b+2ca+b+cp+q=\dfrac{b+2c}{a+b+c}, dividing numerator and denominator on the LHS by a will give.
⇒p+q=ba+2ca1+ba+ca\Rightarrow p+q=\dfrac{\dfrac{b}{a}+\dfrac{2c}{a}}{1+\dfrac{b}{a}+\dfrac{c}{a}}
We know that −(α+β)=ba-\left( \alpha +\beta \right)=\dfrac{b}{a} and αβ=ca\alpha \beta =\dfrac{c}{a}
Substituting in the above equation we get,
⇒p+q=−(α+β)+2αβ1−(α+β)+αβ\Rightarrow p+q=\dfrac{-\left( \alpha +\beta \right)+2\alpha \beta }{1-\left( \alpha +\beta \right)+\alpha \beta }
We will now solve above equation we get,

& \Rightarrow p+q=\dfrac{-\left( \alpha +\beta \right)+2\alpha \beta }{1-\left( \alpha +\beta \right)+\alpha \beta } \\\ & \Rightarrow p+q=\dfrac{\alpha \beta -\alpha +\alpha \beta -\beta }{1-\alpha -\beta +\alpha \beta } \\\ & \Rightarrow p+q=\dfrac{\alpha (\beta -1)+\beta (\alpha -1)}{\left( 1-\alpha \right)-\beta \left( 1-\alpha \right)} \\\ & \Rightarrow p+q=\dfrac{\alpha (\beta -1)+\beta (\alpha -1)}{\left( 1-\beta \right)\left( 1-\alpha \right)} \\\ \end{aligned}$$ We can write the above equation as, $$\Rightarrow p+q=\dfrac{\alpha (\beta -1)+\beta (\alpha -1)}{\left( 1-\beta \right)\left( 1-\alpha \right)}$$ $$\Rightarrow p+q=\dfrac{\alpha (\beta -1)}{\left( 1-\beta \right)\left( 1-\alpha \right)}+\dfrac{\beta (\alpha -1)}{\left( 1-\beta \right)\left( 1-\alpha \right)}$$ Taking – sign common from the numerator, we get, $$\begin{aligned} & \Rightarrow p+q=-\dfrac{\alpha (1-\beta )}{\left( 1-\beta \right)\left( 1-\alpha \right)}-\dfrac{\beta (1-\alpha )}{\left( 1-\beta \right)\left( 1-\alpha \right)} \\\ & \Rightarrow p+q=-\dfrac{\alpha }{\left( 1-\alpha \right)}-\dfrac{\beta }{\left( 1-\beta \right)} \\\ & \Rightarrow p+q=\dfrac{\alpha }{\left( \alpha -1 \right)}+\dfrac{\beta }{\left( \beta -1 \right)}..................(i) \\\ \end{aligned}$$ We have, $pq=\dfrac{c}{a+b+c}$,dividing the numerator and denominator by a, we get $\begin{aligned} & \Rightarrow pq=\dfrac{c}{a+b+c} \\\ & \Rightarrow pq=\dfrac{\dfrac{c}{a}}{1+\dfrac{b}{a}+\dfrac{c}{a}} \\\ \end{aligned}$ We know that $-\left( \alpha +\beta \right)=\dfrac{b}{a}$ and $\alpha \beta =\dfrac{c}{a}$ $$\begin{aligned} & \Rightarrow pq=\dfrac{\dfrac{c}{a}}{1+\dfrac{b}{a}+\dfrac{c}{a}} \\\ & \Rightarrow pq=\dfrac{\alpha \beta }{1-\alpha -\beta +\alpha \beta } \\\ & \Rightarrow pq=\dfrac{\alpha \beta }{(1-\alpha )-\beta (1-\alpha )} \\\ & \Rightarrow pq=\dfrac{\alpha \beta }{(1-\alpha )(1-\beta )} \\\ & \text{Taking - sign common from both terms of denominator,we get,} \\\ & \Rightarrow pq=\dfrac{\alpha \beta }{(\alpha -1)(\beta -1)} \\\ \end{aligned}$$ We can write above expression as, $$\Rightarrow pq=\dfrac{\alpha }{(\alpha -1)}\times \dfrac{\beta }{(\beta -1)}.......................(ii)$$ We will analyze equation (i) and equation(ii), we can say that, $$\begin{aligned} & \Rightarrow p=\dfrac{\alpha }{(\alpha -1)} \\\ & \Rightarrow q=\dfrac{\beta }{(\beta -1)} \\\ \end{aligned}$$ So the answer is d) $\dfrac{\alpha }{\alpha -1},\dfrac{\beta }{\beta -1}$ **Note:** Before attending the questions on quadratic equations the students must know the key formulas like the value sum of roots and product of roots. Before finding the roots of the second equation students should observe the options given, it will help the student to approach the question without any confusion, as in this question we know that answer in the form of $\alpha$ and $\beta$, then only we got the idea to divide with a so that we can get $\alpha$ and $\beta$.