Examine whether the following equation is qua | vedclass

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

Question

(A) Given equation: $x - \frac{1}{x} = 5$Multiply the entire equation by $x$ to eliminate the denominator:$x(x) - x(\frac{1}{x}) = 5(x)$$x^{2} - 1 = 5

Vedclass · Accepted Answer

Yes,it is a quadratic equation.

Vedclass · Answer

(A) Given equation: $x - \frac{1}{x} = 5$Multiply the entire equation by $x$ to eliminate the denominator:$x(x) - x(\frac{1}{x}) = 5(x)$$x^{2} - 1 = 5x$Rearrange the terms to the standard form $ax^{2} + bx + c = 0$:$x^{2} - 5x - 1 = 0$Here,the highest power (degree) of the variable $x$ is $2$.Since the equation is in the form $ax^{2} + bx + c = 0$ where $a 
eq 0$,it is a quadratic equation.

Examine whether the following equation is quadratic or not: $x - \frac{1}{x} = 5$

Similar Questions

Find the roots of the following quadratic equation using the quadratic formula,if they exist: $\frac{1}{x+1} + \frac{2}{x+2} = \frac{4}{x+4}$; $(x \neq -1, -2, -4)$

Is $0.2$ a root of the equation $x^{2}-0.4=0$? Justify.

If the roots of the quadratic equation $3x^{2} + 2\sqrt{5}x - 5 = 0$ exist,find them using the method of completing the square.

Verify whether the given value of $x$ is a solution of the quadratic equation or not: $\frac{x+1}{x-1} - \frac{x-1}{x+1} = \frac{5}{6}, x = 5$

Obtain the roots of the following quadratic equation by using the quadratic formula: $p^{2} x^{2} + (p^{2} - q^{2}) x - q^{2} = 0$

Vedclass Test Series

Exam Paper Generator

Online Exam Module