Question

If one root of the quadratic equation $a{x^2} + bx + c = 0$ is equal to ${n^{th}}$ power of the other, then show that ${(a{c^n})^{\dfrac{1}{{n + 1}}}} + {({a^n}c)^{\dfrac{1}{{n + 1}}}} + b = 0$.

Answer 1

Here we will use the concept that in a quadratic equation the sum of the roots is the ratio of the negative times the coefficient of $x$ to the coefficient of ${x^2}$ and the product of the roots is the ratio of the constant term to the coefficient of ${x^2}$. By applying these concepts we can reach the solution of the question.

Complete answer:
Here, we have the quadratic equation $a{x^2} + bx + c = 0$ where $a,\,\,b$ and $c$ are the constant.
And it is given that one root is equal to the ${n^{th}}$ power of the other.
Let the roots will be $\alpha$ and ${\alpha ^n}$
Now, as we know that in a quadratic equation the sum of the roots is the ratio of the negative times the coefficient of $x$ to the coefficient of ${x^2}$.
So, the sum of the root is,
$\Rightarrow \alpha + {\alpha ^n} = \dfrac{{ - \,coefficient\,\,of\,\,x}}{{coefficient\,\,of\,\,{x^2}}}$
Here, in the given equation the coefficient of ${x^2}$ is $a$, coefficient of $x$ is $b$. So,
$\Rightarrow \alpha + {\alpha ^n} = \dfrac{{ - b}}{a} \ldots \ldots (1)$
Now, as we know that the product of the roots is the ratio of the constant term to the coefficient of ${x^2}$.
So, the product of the root is,
$\Rightarrow \alpha \times {\alpha ^n} = \dfrac{{cons.\,\,term}}{{coefficient\,\,of\,\,{x^2}}}$
Here, in the given equation the coefficient of ${x^2}$ is $a$, constant term is $c$. So,
$\Rightarrow \alpha \times {\alpha ^n} = \dfrac{c}{a}$
As bases are same so the power of $\alpha$ will be added.
$\Rightarrow {\alpha ^{n + 1}} = \dfrac{c}{a}$
Now the above equation can be written as
$\Rightarrow \alpha = {\left( {\dfrac{c}{a}} \right)^{\dfrac{1}{{n + 1}}}} \ldots \ldots (2)$
Substituting the value of $\alpha$ from equation $(2)$ to equation $(1)$. We get,
$\Rightarrow {\left( {\dfrac{c}{a}} \right)^{\dfrac{1}{{n + 1}}}} + {\left( {\dfrac{c}{a}} \right)^{\dfrac{n}{{n + 1}}}} = \dfrac{{ - b}}{a}$
Shifting $a$ to the left side of the equation. We get,
$\Rightarrow a{\left( {\dfrac{c}{a}} \right)^{\dfrac{1}{{n + 1}}}} + a{\left( {\dfrac{c}{a}} \right)^{\dfrac{n}{{n + 1}}}} = - b$
$\Rightarrow a{\left( {\dfrac{c}{a}} \right)^{\dfrac{1}{{n + 1}}}} + a{\left( {\dfrac{c}{a}} \right)^{\dfrac{n}{{n + 1}}}} + b = 0$
We can write the above equation as
$\Rightarrow \dfrac{a}{{{a^{\dfrac{1}{{n + 1}}}}}} \times {c^{\dfrac{1}{{n + 1}}}} + \dfrac{a}{{{a^{\dfrac{n}{{n + 1}}}}}} \times {c^{\dfrac{n}{{n + 1}}}} + b = 0$
Applying the law of exponents i.e., $\dfrac{{{r^n}}}{{{r^m}}} = {r^{n - m}}$ in the equation. We get,
$\Rightarrow {a^{1 - \dfrac{1}{{n + 1}}}} \times {c^{\dfrac{1}{{n + 1}}}} + {a^{1 - \dfrac{n}{{n + 1}}}} \times {c^{\dfrac{n}{{n + 1}}}} + b = 0$
On solving the powers. We get,
$\Rightarrow {a^{\dfrac{n}{{n + 1}}}} \times {c^{\dfrac{1}{{n + 1}}}} + {a^{\dfrac{1}{{n + 1}}}} \times {c^{\dfrac{n}{{n + 1}}}} + b = 0$
The above equation can be written as
$\Rightarrow {({a^n}c)^{\dfrac{1}{{n + 1}}}} + {(a{c^n})^{\dfrac{1}{{n + 1}}}} + b = 0$
Which is the required equation.

Note: Quadratic equations can be defined as the equations that contain at least one term which is squared. This is the reason these equations are called as “quad” meaning square. The general form of the quadratic equation is $a{x^2} + bx + c = 0$ where $a,\,\,b$ and $c$ are the numerical coefficient or constant and the value of $a$ can never be zero. We can find the roots of quadratic equation by various methods such as quadratic formula or factorizing method. Note that the number of roots of quadratic polynomial is $2$ as the highest power in a quadratic polynomial is $2$. Similarly, the number of roots in cubic polynomial is $3$ as the highest power in cubic polynomial is $3$.