Question

If the difference between the roots of $ {{x}^{2}}+ax+b=0 $ is equal to the difference between the roots of $ {{x}^{2}}+px+q=0 $ then $ {{a}^{2}}-{{p}^{2}}= $ \[\]
A. $ b-q $ \[\]
B. $ 2\left( p-q \right) $ \[\]
C. $ 4\left( b-q \right) $ \[\]
D. $ 8\left( b+q \right) $ \[\]

Answer 1

Hint: We recall that the sum of the roots of the general quadratic equation $ a{{x}^{2}}+bx+c=0 $ is given $ \dfrac{-b}{a} $ and the product of the two roots is given by $ \dfrac{c}{a} $ . We denote the roots of $ {{x}^{2}}+ax+b=0 $ as $ \alpha ,\beta $ and of $ {{x}^{2}}+px+q=0 $ as $ \gamma ,\delta $ . We use the given information to have $ \alpha -\beta =\gamma -\delta $ . We square both sides and use the algebraic identity $ {{\left( a+b \right)}^{2}}={{\left( a+b \right)}^{2}}-4ab $ to proceed. \[\]

Complete step by step answer:
We know that the general form of a quadratic equation is given by $ a{{x}^{2}}+bx+c=0 $ where $ a,b,c $ are real numbers and $ a $ cannot be zero. A quadratic equation always has two roots and their sum is given as the negative ratio of the coefficient of $ x $ to coefficient of $ {{x}^{2}} $ . If $ {{x}_{1}},{{x}_{2}} $ x are two roots then their sum is given by
\[{{x}_{1}}+{{x}_{2}}=\dfrac{-b}{a}\]
The product of their roots is given as the ratio of the constant term to the coefficient of $ {{x}^{2}} $ which means
\[{{x}_{1}}{{x}_{2}}=\dfrac{c}{a}\]
We are given two quadratic equation in the question
\[\begin{align}
  & {{x}^{2}}+ax+b=0.....\left( 1 \right) \\
 & {{x}^{2}}+px+q=0.....\left( 2 \right) \\
\end{align}\]
Let us denote the roots of equation (1) as $ \alpha ,\beta $ and denote the roots of equation (2) as $ \gamma ,\delta $ . Let us find the sum of products of roots of equation (1) . We have
\[\begin{align}
  & \alpha +\beta =\dfrac{-a}{1}=-a.......\left( 3 \right) \\
 & \alpha \beta =\dfrac{b}{1}=b..................\left( 4 \right) \\
\end{align}\]
We find the sum and product of roots of equation (2) and have;
\[\begin{align}
  & \gamma +\delta =\dfrac{-p}{1}=-p.........\left( 5 \right) \\
 & \gamma \delta =\dfrac{q}{1}=q................\left( 6 \right) \\
\end{align}\]
We are given in the question that the difference between the roots of equation (1) and equation (2) is equal. So we have;
\[\alpha -\beta =\gamma -\delta \]
We square both sides of the above equation to have;
\[\Rightarrow {{\left( \alpha -\beta \right)}^{2}}={{\left( \gamma -\delta \right)}^{2}}\]
We use the algebraic identity $ {{\left( a+b \right)}^{2}}={{\left( a+b \right)}^{2}}-4ab $ in both sides of the above step to have
\[\Rightarrow {{\left( \alpha +\beta \right)}^{2}}-4\alpha \beta ={{\left( \gamma -\delta \right)}^{2}}-4\gamma \delta \]
We put the values obtained from (3), (4),(5) and (6) to have;
\[\begin{align}
  & \Rightarrow {{\left( -a \right)}^{2}}-4b={{\left( -p \right)}^{2}}-4q \\
 & \Rightarrow {{a}^{2}}-4b={{p}^{2}}-4q \\
 & \Rightarrow {{a}^{2}}-{{p}^{2}}=4b-4q \\
 & \therefore {{a}^{2}}-{{p}^{2}}=4\left( b-q \right) \\
\end{align}\]
So the correct option is C. \[\]

Note:
 We need to carefully observe the coefficient of $ {{x}^{2}} $ , coefficient of $ x $ , and the constant term to find the sum and product of two roots. The roots are going to be equal when discriminant $ D={{b}^{2}}-4ac=0 $ . We can find the roots of a quadratic equation either by splitting the middle term or by quadratic formula $ x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} $ .