The roots of frac{x+4}{x-4}+frac{x-4}{x+4}=fr | vedclass

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

Question

(C) Let $y = \frac{x+4}{x-4}$. Then the equation becomes $y + \frac{1}{y} = \frac{10}{3}$.Multiplying by $3y$,we get $3y^2 + 3 = 10y$,which simplifies

Vedclass · Accepted Answer

$\pm 8$

Vedclass · Answer

(C) Let $y = \frac{x+4}{x-4}$. Then the equation becomes $y + \frac{1}{y} = \frac{10}{3}$.Multiplying by $3y$,we get $3y^2 + 3 = 10y$,which simplifies to $3y^2 - 10y + 3 = 0$.Factoring the quadratic equation: $3y^2 - 9y - y + 3 = 0 \Rightarrow 3y(y-3) - 1(y-3) = 0 \Rightarrow (3y-1)(y-3) = 0$.Thus,$y = 3$ or $y = \frac{1}{3}$.Case $1$: $\frac{x+4}{x-4} = 3 \Rightarrow x+4 = 3x-12 \Rightarrow 2x = 16 \Rightarrow x = 8$.Case $2$: $\frac{x+4}{x-4} = \frac{1}{3} \Rightarrow 3x+12 = x-4 \Rightarrow 2x = -16 \Rightarrow x = -8$.Therefore,the roots are $x = 8, -8$ or $\pm 8$.

The roots of $\frac{x+4}{x-4}+\frac{x-4}{x+4}=\frac{10}{3}$ are:

Similar Questions

Solve the given two equations and select the correct answer from the given options.
$I.$ $\sqrt{1225} x + \sqrt{4900} = 0$
$II.$ $(81)^{1/4} y + (343)^{1/3} = 0$

Solve the given two equations and select the correct answer from the given options.
$I.$ $2x^{2} + x - 1 = 0$
$II.$ $2y^{2} - 3y + 1 = 0$

The value of $p$ for which one root of the equation ${x^2} - 30x + p = 0$ is the square of the other,are

Solve the given two equations and select the correct option.
$I.$ $x^{2}+5x+6=0$
$II.$ $y^{2}+3y+2=0$

The number of integral values of $m$ for which the quadratic expression $(1 + 2m)x^2 - 2(1 + 3m)x + 4(1 + m)$ is always positive for all $x \in R$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module