Question

Let a,b,c,d be distinct real numbers and a and b are the roots of the quadratic equation ${{x}^{2}}-2cx-5d=0$. If c and d are the roots of the quadratic equation ${{x}^{2}}-2ax-5b=0$, then find the numerical value of $a+b+c+d$
$\begin{align}
  & a)30 \\
 & b)15 \\
 & c)10 \\
 & d)60 \\
\end{align}$

Answer 1

Hint: Assume the given quadratic equation as equation (i) and equation (ii). Consider these equations one by one and apply the identities: - \[\text{Sum of roots =}\dfrac{\text{-coefficient}\,\text{of}\,}{\text{coefficient}\,\text{of}\,{{\text{x}}^{\text{2}}}}\]And \[\text{products of roots=}\dfrac{\text{constant}\,\text{terms}}{\text{coefficient}\,\text{of}\,{{\text{x}}^{\text{2}}}}\]form relationship between the roots a,b,c and d and hence find the value of $a+b+c+d$

Complete step by step answer:
Here, we have been provided with two quadratic equations: ${{x}^{2}}-2cx-5d=0$ whose roots are ‘a’ and ‘b’ and ${{x}^{2}}-2ax-5b=0$ whose roots are ‘c’ and ‘d’. we have been asked to determine the value of the sum of these roots. i.e. $a+b+c+d$. First, we need to know the relation between these roots.
Now, we have two quadratic equations: -
$\begin{align}
  & {{x}^{2}}-2cx-5d=0.......(i) \\
 & {{x}^{2}}-2ax-5b=0.......(ii) \\
\end{align}$
1. considering equation (i) we have,
${{x}^{2}}-2cx-5d=0$
Since the roots of this quadratic equation are ‘a’ and ‘b’ therefore applying the identities of sum of roots and products of roots, we get,
\[\text{Sum of roots =}\dfrac{\text{-coefficient}\,\text{of}\,}{\text{coefficient}\,\text{of}\,{{\text{x}}^{\text{2}}}}\]
$\begin{align}
  & \Rightarrow a+b=\dfrac{-(-2c)}{1} \\
 & \Rightarrow a+b=2c..........(i) \\
\end{align}$
Now, we know that,
\[\text{products of roots=}\dfrac{\text{constant}\,\text{terms}}{\text{coefficient}\,\text{of}\,{{\text{x}}^{\text{2}}}}\]

$\begin{align}
  & \Rightarrow ab=-5d \\
 & \Rightarrow ab=-5d................(ii) \\
\end{align}$
2. considering equation (iii) we have
${{x}^{2}}-2ax-5b=0$
Since, the roots of this quadratic equation are ‘c’ and ‘d’ therefore applying identities of sum of roots and products of roots, we get
\[\text{Sum of roots =}\dfrac{\text{-coefficient}\,\text{of}\,}{\text{coefficient}\,\text{of}\,{{\text{x}}^{\text{2}}}}\]
$\begin{align}
  & \Rightarrow c+d=\dfrac{=(-2a)}{1} \\
 & \Rightarrow c+d=2a.......(iii) \\
\end{align}$
Now, we know that,
\[\text{products of roots=}\dfrac{\text{constant}\,\text{terms}}{\text{coefficient}\,\text{of}\,{{\text{x}}^{\text{2}}}}\]
$\begin{align}
  & \Rightarrow cd=\dfrac{-5b}{1} \\
 & \Rightarrow cd=-5b........(iv) \\
\end{align}$
Now adding equation (i) and (iii)we get,
$\begin{align}
  & \Rightarrow a+b+c+d=2c+2a \\
 & \Rightarrow a+b+c+d=2(a+c)........(5) \\
\end{align}$
Considering equation (2) and replacing ‘b’ with terms of ‘a’ and ‘c’ using equation (i), we get,
$\begin{align}
  & \Rightarrow a(2c-a)=-5d \\
 & \Rightarrow 2ac-{{a}^{2}}=-5d.........(vi) \\
\end{align}$
Considering equation (iv) and replacing ‘d’ with terms of ‘a’ and ‘c’ using equation (iii), we get,
$\begin{align}
  & \Rightarrow c(2a-c)=-5b \\
 & \Rightarrow 2ac-{{c}^{2}}=-5b.........(vii) \\
\end{align}$
Subtracting equation (vii) from equation (vi), we get,
$\begin{align}
  & \Rightarrow {{c}^{2}}-{{a}^{2}}=-5d-(-5b) \\
 & \Rightarrow {{c}^{2}}-{{a}^{2}}=5b-5d \\
 & \Rightarrow {{c}^{2}}-{{a}^{2}}=5(b-d) \\
\end{align}$
Applying the identity ${{x}^{2}}-{{y}^{2}}=(x+y)(x-y)$ in the LHS, we get
$\Rightarrow (c+a)(c-a)=5(b-d).......(viii)$
Now subtracting equation (iii) from equation (i), we get,
$\begin{align}
  & \Rightarrow (a-c)+(b-d)=2(c-a) \\
 & \Rightarrow (b-d)=2(c-a)-(a-c) \\
 & \Rightarrow (b-d)=2(c-a)+(c-a) \\
 & \Rightarrow (b-d)=3(c-a) \\
\end{align}$
Substituting the value of $(b-d)$ from the above relation in equation (viii), we get,
$\begin{align}
  & \Rightarrow (c+a)+(c-a)=5\times 3(c-a) \\
 & \Rightarrow (c+a)(c-a)-15(c-a) \\
 & \Rightarrow (c+a)(c-a)-15(c-a)=0 \\
 & \Rightarrow (c-a)[(c+a)-15]=0 \\
\end{align}$
Substituting each term equal to 0 we get,
$\begin{align}
  & \Rightarrow c-a=0\,or(c+a)-15=0 \\
 & \Rightarrow c=a\,or\,(c+a)=15 \\
\end{align}$
Now, in the question it is clearly given to us that a,b,c,d are distinct numbers, so, a and c cannot be equal. Therefore rejecting \[c=a\], we have,
$\Rightarrow (a+c)=15$
Substituting $(a+c)=15$ in equation 5, we get
$\begin{align}
  & \Rightarrow a+b+c+d=2\times 15 \\
 & \Rightarrow a+b+c+d=30 \\
\end{align}$

So, the correct answer is “Option a”.

Note: You may note that from equation (5) it is clear to us that actually ‘a’ and ‘c’ are the main terms which will lead us to our answer. So, we have replaced ‘b’ and ‘d’ everywhere with our terms of ‘a’ and ‘c’ so that whatever equation are formed will be in terms of ‘a’ and ‘c’. remember that you must not try to find the numerical values of each roots a,b,c and d as it will be a very lengthy process. Remember the formulas of sum and product of roots to solve the question. You must read the question carefully to reject the incorrect conditions obtained, like a=c in the last part of the solution.