Question

If \[\alpha ,\beta \] are roots of \[a{{x}^{2}}+2bx+c=0\] and \[\alpha +\delta ,\beta +\delta \] are the roots of \[A{{x}^{2}}+2Bx+C=0\] , then \[\dfrac{{{b}^{2}}-ac}{{{B}^{2}}-AC}=\]
(a) \[\dfrac{a}{A}\]
(b) \[\dfrac{A}{a}\]
(c) \[{{\left( \dfrac{a}{A} \right)}^{2}}\]
(d) \[{{\left( \dfrac{A}{a} \right)}^{2}}\]

Answer 1

Hint: First, we should know the sum of the roots which is given as \[\dfrac{-b}{a}\] and products of roots as \[\dfrac{c}{a}\] . Then we will find this for the equation \[a{{x}^{2}}+2bx+c=0\] and \[A{{x}^{2}}+2Bx+C=0\] . Then we will subtract the roots and will do squaring on both sides like \[{{\left( \alpha -\beta \right)}^{2}}={{\alpha }^{2}}+{{\beta }^{2}}-2\alpha \beta \] and \[{{\left( \left( \alpha +\delta \right)-\left( \beta +\delta \right) \right)}^{2}}={{\left( \alpha +\delta \right)}^{2}}+{{\left( \beta +\delta \right)}^{2}}-2\left( \alpha +\delta \right)\left( \beta +\delta \right)\] . After this we will make it in perfect square of \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\] by adding and subtracting some terms. Then we will substitute the values into the equation and from the equation we will make subject variable as \[\left( {{b}^{2}}-ac \right)\] and \[\left( {{B}^{2}}-AC \right)\] . Then dividing the equation i.e. \[\left( {{B}^{2}}-AC \right)\] by \[\left( {{b}^{2}}-ac \right)\] we will get the required answer.

Complete step-by-step answer:
Here, we are given that \[\alpha ,\beta \] are roots of \[a{{x}^{2}}+2bx+c=0\] . So, we will find the sum of the roots which is given as \[\dfrac{-b}{a}\] and products of roots as \[\dfrac{c}{a}\] .
So, here we can write it as
Sum of the roots \[\alpha +\beta =\dfrac{-2b}{a}\] …………………(1)
Product of roots \[\alpha \beta =\dfrac{c}{a}\] …………………….(2)
Now, we will do squaring of the sum of the roots. So, we get \[{{\left( \alpha +\beta \right)}^{2}}\] . To expand this, we will use the formula \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\] .
So, we can write it as
\[{{\left( \alpha +\beta \right)}^{2}}={{\alpha }^{2}}+{{\beta }^{2}}+2\alpha \beta \] …………………(3)
Now, if we subtract the roots and do squaring as above, we will get as
\[{{\left( \alpha -\beta \right)}^{2}}={{\alpha }^{2}}+{{\beta }^{2}}-2\alpha \beta \]
Now, we will add and subtract \[2\alpha \beta \] in the above equation to make it a perfect square. So, we get as
\[{{\left( \alpha -\beta \right)}^{2}}={{\alpha }^{2}}+{{\beta }^{2}}+2\alpha \beta -2\alpha \beta -2\alpha \beta \]
Thus, from equation (3), we can write it as
\[{{\left( \alpha -\beta \right)}^{2}}={{\left( \alpha +\beta \right)}^{2}}-4\alpha \beta \] …………………..(4)
Now, we will substitute the values of equation (1) and (2) in equation (4). So, we get as
\[{{\left( \alpha -\beta \right)}^{2}}={{\left( \dfrac{-2b}{a} \right)}^{2}}-4\left( \dfrac{c}{a} \right)\]
On simplification, we get as
\[{{\left( \alpha -\beta \right)}^{2}}=\dfrac{4{{b}^{2}}}{{{a}^{2}}}-\dfrac{4c}{a}\]
On taking LCM of \[{{a}^{2}}\] , we can write it as
\[{{\left( \alpha -\beta \right)}^{2}}=\dfrac{4{{b}^{2}}-4ac}{{{a}^{2}}}\]
On taking 4 common, we get as
\[{{\left( \alpha -\beta \right)}^{2}}=\dfrac{4\left( {{b}^{2}}-ac \right)}{{{a}^{2}}}\]
We will make \[\left( {{b}^{2}}-ac \right)\] as subject variable.
\[\left( {{b}^{2}}-ac \right)=\dfrac{{{a}^{2}}{{\left( \alpha -\beta \right)}^{2}}}{4}\] ……………………..(5)
Similarly, we are given that \[\alpha +\delta ,\beta +\delta \] are the roots of \[A{{x}^{2}}+2Bx+C=0\] . So, we get as done above that the sum of the roots which is given as \[\dfrac{-b}{a}\] and products of roots as \[\dfrac{c}{a}\] .
So, we can write it as
Sum of the roots \[\alpha +\delta +\beta +\delta =\dfrac{-2B}{A}\] …………………(6)
Product of roots \[\left( \alpha +\delta \right)\left( \beta +\delta \right)=\dfrac{C}{A}\] ………………..(7)
Now, if we subtract the roots and do squaring, we will get as
\[{{\left( \left( \alpha +\delta \right)-\left( \beta +\delta \right) \right)}^{2}}={{\left( \alpha +\delta \right)}^{2}}+{{\left( \beta +\delta \right)}^{2}}-2\left( \alpha +\delta \right)\left( \beta +\delta \right)\]
Now, we will add and subtract \[2\left( \alpha +\delta \right)\left( \beta +\delta \right)\] in the above equation to make it a perfect square. So, we get as
\[\begin{align}
  & {{\left( \left( \alpha +\delta \right)-\left( \beta +\delta \right) \right)}^{2}}={{\left( \alpha +\delta \right)}^{2}}+{{\left( \beta +\delta \right)}^{2}}+2\left( \alpha +\delta \right)\left( \beta +\delta \right)-2\left( \alpha +\delta \right)\left( \beta +\delta \right) \\
 & \text{ }-2\left( \alpha +\delta \right)\left( \beta +\delta \right) \\
\end{align}\]
Thus using \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\] , we can write it as
\[{{\left( \left( \alpha +\delta \right)-\left( \beta +\delta \right) \right)}^{2}}={{\left( \left( \alpha +\delta \right)+\left( \beta +\delta \right) \right)}^{2}}-4\left( \alpha +\delta \right)\left( \beta +\delta \right)\]
Now, we will substitute the values from equation (6), (7) in above equation so, we get as
 \[{{\left( \left( \alpha +\delta \right)-\left( \beta +\delta \right) \right)}^{2}}={{\left( \dfrac{-2B}{A} \right)}^{2}}-\dfrac{4C}{A}\]
On taking LCM and further solving we get as
\[{{\left( \left( \alpha +\delta \right)-\left( \beta +\delta \right) \right)}^{2}}={{\dfrac{4B}{{{A}^{2}}}}^{2}}-\dfrac{4C}{A}=\dfrac{4{{B}^{2}}-4AC}{{{A}^{2}}}\]
Taking 4 common from numerator, we get as
\[{{\left( \left( \alpha +\delta \right)-\left( \beta +\delta \right) \right)}^{2}}=\dfrac{4\left( {{B}^{2}}-AC \right)}{{{A}^{2}}}\]
We will make the subject variable \[\left( {{B}^{2}}-AC \right)\] . We get as
\[\left( {{B}^{2}}-AC \right)=\dfrac{{{A}^{2}}{{\left( \left( \alpha +\delta \right)-\left( \beta +\delta \right) \right)}^{2}}}{4}\] ……………………….(8)
Now, we will divide equation (5) by (8), we get as
\[\dfrac{\left( {{b}^{2}}-ac \right)}{\left( {{B}^{2}}-AC \right)}=\dfrac{\dfrac{{{a}^{2}}{{\left( \alpha -\beta \right)}^{2}}}{4}}{\dfrac{{{A}^{2}}{{\left( \left( \alpha +\delta \right)-\left( \beta +\delta \right) \right)}^{2}}}{4}}\]
On cancelling denominator terms, and further simplifying the terms we get as
\[\dfrac{\left( {{b}^{2}}-ac \right)}{\left( {{B}^{2}}-AC \right)}=\dfrac{{{a}^{2}}{{\left( \alpha -\beta \right)}^{2}}}{{{A}^{2}}{{\left( \alpha +\delta -\beta -\delta \right)}^{2}}}\]
\[\dfrac{\left( {{b}^{2}}-ac \right)}{\left( {{B}^{2}}-AC \right)}=\dfrac{{{a}^{2}}{{\left( \alpha -\beta \right)}^{2}}}{{{A}^{2}}{{\left( \alpha -\beta \right)}^{2}}}\]
Further we get as
\[\dfrac{\left( {{b}^{2}}-ac \right)}{\left( {{B}^{2}}-AC \right)}=\dfrac{{{a}^{2}}}{{{A}^{2}}}={{\left( \dfrac{a}{A} \right)}^{2}}\]
Thus, option (c) is the correct answer.

Note: Students should know that in this type of problem we have to somehow manage to get equation in terms \[\left( {{B}^{2}}-AC \right)\] and \[\left( {{b}^{2}}-ac \right)\] in order to get the value as asked in question. Students should also strike that sum and products of root we have to find and then on squaring we will get the answer. So, this type of problem is generally more of logic and less of directly applying formulas. Though formulas are also important to know, only the answer will be correct, so students should apply their logic and then try to solve it.