Examine whether the following equation is qua | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given equation: $x + \frac{1}{x} = x^2$ $(x 
eq 0)$.Multiply the entire equation by $x$ to eliminate the fraction:$x(x) + x(\frac{1}{x}) = x(x^2)

Vedclass · Accepted Answer

No,it is not a quadratic equation.

Vedclass · Answer

(B) Given equation: $x + \frac{1}{x} = x^2$ $(x 
eq 0)$.Multiply the entire equation by $x$ to eliminate the fraction:$x(x) + x(\frac{1}{x}) = x(x^2)$$x^2 + 1 = x^3$Rearranging the terms to one side:$x^3 - x^2 - 1 = 0$.$A$ quadratic equation is defined as an equation of the form $ax^2 + bx + c = 0$,where $a 
eq 0$. The highest power (degree) of the variable $x$ in the given equation is $3$. Since the degree of the equation is $3$,it is a cubic equation,not a quadratic equation.

Examine whether the following equation is quadratic or not: $x + \frac{1}{x} = x^2$ $(x \neq 0)$

Similar Questions

If $x=-2$ is a solution of a quadratic equation $k x^{2}+5 x+2=0,$ then $k=\ldots \ldots \ldots \ldots$

Find the roots of the quadratic equation by using the quadratic formula:
$-3x^{2} + 5x + 12 = 0$

Find the roots of $6x^{2} - \sqrt{2}x - 2 = 0$ by the factorisation of the corresponding quadratic polynomial.

If $81$ is the discriminant of $2x^{2} + 5x - k = 0$,then the value of $k$ is ..............

Solve the following equation using the quadratic formula,if the equation has a solution in $R$: $x + \frac{1}{x} = 3, x \neq 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module