Which of the following are quadratic equations?
$\left( x\ +\ \frac{1}{x}\right)^{2} \ =\ 3\left( x\ +\ \frac{1}{x}\right) \ +\ 4$

Given:

Given equation is $\left( x\ +\ \frac{1}{x}\right)^{2} \ =\ 3\left( x\ +\ \frac{1}{x}\right) \ +\ 4$.

To do:

We have to check whether the given equation is quadratic.

Solution:

The standard form of a quadratic equation is $ax^2+bx+c=0$.

$\left( x\ +\ \frac{1}{x}\right)^{2} \ =\ 3\left( x\ +\ \frac{1}{x}\right) \ +\ 4$

$x^2+\frac{1}{x^2}+2=3x+\frac{3}{x}+4$

$x^2(x^2)+x^2(\frac{1}{x^2})+x^2(2) = x^2(3x)+x^2(\frac{3}{x})+x^2(4)$

$x^4+1+2x^2=3x^3+3x+4x^2$

$x^4-3x^3-2x^2-3x+1=0$

The equation $x^4-3x^3-2x^2-3x+1=0$ is not of the form $ax^2+bx+c=0$ as its degree is $4$.

Therefore, $\left( x\ +\ \frac{1}{x}\right)^{2} \ =\ 3\left( x\ +\ \frac{1}{x}\right) \ +\ 4$ is not a quadratic equation.

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Updated on: 10-Oct-2022

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