Questions & Answers

Question

Answers

A.3

B.2

C.1

D.4

Answer
Verified

The given quadratic equations are $a{x^2} + bx + c = 0$ and $b{x^2} + cx + a = 0$.

We need to find the value of $\dfrac{{{a^3} + {b^3} + {c^3}}}{{abc}}$.

Let,

$a{x^2} + bx + c = 0$ ………(i)

$b{x^2} + cx + a = 0$ …..….(ii)

Let $\alpha $ and $\beta $ be the roots common for the above equation.

The sum of the roots is given by $ - \dfrac{{{\text{coefficient}}\,{\text{of}}\,{\text{x}}}}{{{\text{coefficient}}\,{\text{of}}\,{{\text{x}}^2}}}$ and the product of roots is given by $ - \dfrac{{{\text{constant}}}}{{{\text{coefficient}}\,{\text{of}}\,{{\text{x}}^2}}}$.

Therefore,

For eq. (i), $\alpha + \beta = \dfrac{{ - b}}{a}$ and $\alpha \beta = \dfrac{c}{a}$.

For eq. (ii), $\alpha + \beta = \dfrac{{ - c}}{b}$ and $\alpha \beta = \dfrac{a}{b}$.

As both the roots are common, the sum and the product of the root will be equal for both the equations.

Therefore, it can be interpreted as follows:

\[\dfrac{{ - b}}{a} = \dfrac{{ - c}}{b} \Rightarrow {b^2} = ac\] …….(iii)

\[ \Rightarrow c = \dfrac{{{b^2}}}{a}\] ……(iv)

\[ \Rightarrow a = \dfrac{{{b^2}}}{c}\] ……(v)

Again,

\[\dfrac{c}{a} = \dfrac{a}{b} \Rightarrow {a^2} = bc\] ……(vi)

Now, use equation (iv) in equation (vi).

$

{a^2} = bc \\

{a^2} = b\left( {\dfrac{{{b^2}}}{a}} \right)\,\,\,\,\left[ {\because c = \dfrac{{{b^2}}}{a}} \right] \\

{a^2} = \dfrac{{{b^3}}}{a} \\

{a^2}\left( a \right) = {b^3} \\

{a^3} = {b^3} \\

$

Now, use equation (v) in equation (vi).

$

{a^2} = bc \\

{\left( {\dfrac{{{b^2}}}{c}} \right)^2} = bc\,\,\,\left[ {\because a = \dfrac{{{b^2}}}{c}} \right] \\

\dfrac{{{b^4}}}{{{c^2}}} = bc \\

\dfrac{{{b^4}}}{b} = c\left( {{c^2}} \right) \\

{b^3} = {c^3} \\

$

So, it can be observed that,

${a^3} = {b^3} = {c^3} \Rightarrow a = b = c$.

Now, use these values to find the value of $\dfrac{{{a^3} + {b^3} + {c^3}}}{{abc}}$.

$\dfrac{{{a^3} + {b^3} + {c^3}}}{{abc}} = \dfrac{{{a^3} + {a^3} + {a^3}}}{{a \cdot a \cdot a}} = \dfrac{{3{a^3}}}{{{a^3}}} = 3$

Here is the case of two quadratic having same roots and to solve the equation let it be equal to some constant: \[\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}} = \lambda \left( {{\text{say}}} \right)\].

Here, ${a_1}$ and ${a_2}$ are the coefficient of ${x^2}$, ${b_1}$ and ${b_2}$ are the coefficient of $x$ and ${c_1}$ and ${c_2}$ are the constants of the equation.