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Question: If \[\alpha ,\beta \] are roots of \[a{{x}^{2}}+2bx+c=0\] and \[\alpha +\delta ,\beta +\delta \] are...

If α,β\alpha ,\beta are roots of ax2+2bx+c=0a{{x}^{2}}+2bx+c=0 and α+δ,β+δ\alpha +\delta ,\beta +\delta are the roots of Ax2+2Bx+C=0A{{x}^{2}}+2Bx+C=0 , then b2−acB2−AC=\dfrac{{{b}^{2}}-ac}{{{B}^{2}}-AC}=
(a) aA\dfrac{a}{A}
(b) Aa\dfrac{A}{a}
(c) (aA)2{{\left( \dfrac{a}{A} \right)}^{2}}
(d) (Aa)2{{\left( \dfrac{A}{a} \right)}^{2}}

Explanation

Solution

First, we should know the sum of the roots which is given as −ba\dfrac{-b}{a} and products of roots as ca\dfrac{c}{a} . Then we will find this for the equation ax2+2bx+c=0a{{x}^{2}}+2bx+c=0 and Ax2+2Bx+C=0A{{x}^{2}}+2Bx+C=0 . Then we will subtract the roots and will do squaring on both sides like (α−β)2=α2+β2−2αβ{{\left( \alpha -\beta \right)}^{2}}={{\alpha }^{2}}+{{\beta }^{2}}-2\alpha \beta and ((α+δ)−(β+δ))2=(α+δ)2+(β+δ)2−2(α+δ)(β+δ){{\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 (a+b)2=a2+b2+2ab{{\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 (b2−ac)\left( {{b}^{2}}-ac \right) and (B2−AC)\left( {{B}^{2}}-AC \right) . Then dividing the equation i.e. (B2−AC)\left( {{B}^{2}}-AC \right) by (b2−ac)\left( {{b}^{2}}-ac \right) we will get the required answer.

Complete step-by-step answer :
Here, we are given that α,β\alpha ,\beta are roots of ax2+2bx+c=0a{{x}^{2}}+2bx+c=0 . So, we will find the sum of the roots which is given as −ba\dfrac{-b}{a} and products of roots as ca\dfrac{c}{a} .
So, here we can write it as
Sum of the roots α+β=−2ba\alpha +\beta =\dfrac{-2b}{a} …………………(1)
Product of roots αβ=ca\alpha \beta =\dfrac{c}{a} …………………….(2)
Now, we will do squaring of the sum of the roots. So, we get (α+β)2{{\left( \alpha +\beta \right)}^{2}} . To expand this, we will use the formula (a+b)2=a2+b2+2ab{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab .
So, we can write it as
(α+β)2=α2+β2+2αβ{{\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
(α−β)2=α2+β2−2αβ{{\left( \alpha -\beta \right)}^{2}}={{\alpha }^{2}}+{{\beta }^{2}}-2\alpha \beta
Now, we will add and subtract 2αβ2\alpha \beta in the above equation to make it a perfect square. So, we get as
(α−β)2=α2+β2+2αβ−2αβ−2αβ{{\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
(α−β)2=(α+β)2−4αβ{{\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
(α−β)2=(−2ba)2−4(ca){{\left( \alpha -\beta \right)}^{2}}={{\left( \dfrac{-2b}{a} \right)}^{2}}-4\left( \dfrac{c}{a} \right)
On simplification, we get as
(α−β)2=4b2a2−4ca{{\left( \alpha -\beta \right)}^{2}}=\dfrac{4{{b}^{2}}}{{{a}^{2}}}-\dfrac{4c}{a}
On taking LCM of a2{{a}^{2}} , we can write it as
(α−β)2=4b2−4aca2{{\left( \alpha -\beta \right)}^{2}}=\dfrac{4{{b}^{2}}-4ac}{{{a}^{2}}}
On taking 4 common, we get as
(α−β)2=4(b2−ac)a2{{\left( \alpha -\beta \right)}^{2}}=\dfrac{4\left( {{b}^{2}}-ac \right)}{{{a}^{2}}}
We will make (b2−ac)\left( {{b}^{2}}-ac \right) as subject variable.
(b2−ac)=a2(α−β)24\left( {{b}^{2}}-ac \right)=\dfrac{{{a}^{2}}{{\left( \alpha -\beta \right)}^{2}}}{4} ……………………..(5)
Similarly, we are given that α+δ,β+δ\alpha +\delta ,\beta +\delta are the roots of Ax2+2Bx+C=0A{{x}^{2}}+2Bx+C=0 . So, we get as done above that the sum of the roots which is given as −ba\dfrac{-b}{a} and products of roots as ca\dfrac{c}{a} .
So, we can write it as
Sum of the roots α+δ+β+δ=−2BA\alpha +\delta +\beta +\delta =\dfrac{-2B}{A} …………………(6)
Product of roots (α+δ)(β+δ)=CA\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
((α+δ)−(β+δ))2=(α+δ)2+(β+δ)2−2(α+δ)(β+δ){{\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(α+δ)(β+δ)2\left( \alpha +\delta \right)\left( \beta +\delta \right) in the above equation to make it a perfect square. So, we 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)-2\left( \alpha +\delta \right)\left( \beta +\delta \right) \\\ & \text{ }-2\left( \alpha +\delta \right)\left( \beta +\delta \right) \\\ \end{aligned}$$ Thus using $${{\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.