If one root of the quadratic equation...

MathematicsRelation Between Roots and CoefficientsJEE Advanced 1983Easy

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

Visualized Solution (Hindi)

Defining the Roots

Let the roots of the quadratic equation $a x^{2} + b x + c = 0$ be $α$ and $β$ .
According to the problem, one root is the $n$ -th power of the other.
Therefore, we can set $β = α^{n}$ .

Product of Roots Formula

Recall the property for the product of roots in a quadratic equation $a x^{2} + b x + c = 0$ .
Product of roots $= \frac{c}{a}$
In our case: $α \cdot α^{n} = \frac{c}{a}$

Simplifying the Product

Using the laws of exponents: $α^{1} \cdot α^{n} = α^{n + 1}$
So, the equation becomes: $α^{n + 1} = \frac{c}{a}$

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

To isolate $α$ , take the $(n + 1)$ -th root on both sides.
$α = (\frac{c}{a})^{\frac{1}{n + 1}}$

Sum of Roots Formula

Recall the property for the sum of roots.
Sum of roots $= - \frac{b}{a}$
In our case: $α + α^{n} = - \frac{b}{a}$

Substitution of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Substitute the value of $α = (\frac{c}{a})^{\frac{1}{n + 1}}$ into the sum equation.
$(\frac{c}{a})^{\frac{1}{n + 1}} + ((\frac{c}{a})^{\frac{1}{n + 1}})^{n} = - \frac{b}{a}$
Simplify the second term: $(\frac{c}{a})^{\frac{1}{n + 1}} + (\frac{c}{a})^{\frac{n}{n + 1}} = - \frac{b}{a}$

Rearranging the Equation

Multiply the entire equation by $a$ to clear the denominator.
$a \cdot (\frac{c}{a})^{\frac{1}{n + 1}} + a \cdot (\frac{c}{a})^{\frac{n}{n + 1}} = - b$
Rearrange to bring $b$ to the left side: $a \cdot (\frac{c}{a})^{\frac{1}{n + 1}} + a \cdot (\frac{c}{a})^{\frac{n}{n + 1}} + b = 0$

Manipulating the First Term

Focus on the term $a \cdot (\frac{c}{a})^{\frac{1}{n + 1}}$ .
Write $a$ as $(a^{n + 1})^{\frac{1}{n + 1}}$ .
$(a^{n + 1})^{\frac{1}{n + 1}} \cdot (\frac{c}{a})^{\frac{1}{n + 1}} = (a^{n + 1} \cdot \frac{c}{a})^{\frac{1}{n + 1}} = (a^{n} c)^{\frac{1}{n + 1}}$

Manipulating the Second Term

Focus on the term $a \cdot (\frac{c}{a})^{\frac{n}{n + 1}}$ .
Write $a$ as $(a^{n + 1})^{\frac{1}{n + 1}}$ and $(\frac{c}{a})^{\frac{n}{n + 1}}$ as $(\frac{c ^{n}}{a ^{n}})^{\frac{1}{n + 1}}$ .
$(a^{n + 1})^{\frac{1}{n + 1}} \cdot (\frac{c ^{n}}{a ^{n}})^{\frac{1}{n + 1}} = (a^{n + 1} \cdot \frac{c ^{n}}{a ^{n}})^{\frac{1}{n + 1}} = (a c^{n})^{\frac{1}{n + 1}}$

Final Result

Substitute the simplified terms back into the equation.
$(a^{n} c)^{\frac{1}{n + 1}} + (a c^{n})^{\frac{1}{n + 1}} + b = 0$
Hence Proved.

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