Question

solve the equation:- x/x+1 + x+1/x=34/15

Answer 1

The values of “x” from the given equation are $\bold{\frac{3}{2},-\frac{5}{2}}$.

Given:

$\frac{x}{x+1}+\frac{x+1}{x}=\frac{34}{15}$

For solving the given equation,  

$x^{2}+\left(\frac{x+1 )^{2}}{x(x+1)}\right)=\frac{34}{15}$

$\frac{x^{2}+2 x^{2}+1+2 x}{x^{2}+x}=\frac{34}{15}$

$15\left(2 x^{2}+1+2 x\right)=34\left(x^{2}+x\right)$

$30 x^{2}+15+30 x=34 x^{2}+34 x$

$4 x^{2}+4 x-15=0$

To find the roots in the quadratic equation which is in the form of $a x^{2}+b x-c=0$  

By using the below formula, we can find the roots

$=-b \pm \frac{\left(\sqrt{b^{2}+4 a c}\right)}{2 a},$

By solving the equation to find the roots are

$X=-4 \pm \frac{\left(\sqrt{\left(4^{2}+(4 \times 4 \times 15)\right.}\right)}{2 \times 4}$

$=-4 \pm \frac{(\sqrt{(16+240)})}{8}$

$=-4 \pm \frac{\sqrt{256}}{8}$

$=-4 \pm \frac{16}{8}$

$=-4+\frac{16}{8},-4-\frac{16}{8}$

There are two roots totally, one is positive and another one is positive.

$\begin{array}{c}{=\frac{12}{8},-\frac{20}{8}} \\ \\{=\frac{3}{2},-\frac{5}{2}}\end{array}$

Answer 2

Answer:

Step-by-step explanation:

Solution :-

We have,

x/(x + 1) + (x + 1)/x = 34/15

By solving this, we get

x² + (x + 1)²/x(x + 1) = 34/15

x² + x² + 2x + 1/x² + x = 34/15

34x² + 34x = 15x² + 15x² + 30x + 15

4x² + 4x - 15 = 0

By solving in factorization method, we get

4x² + 10x - 6x - 15 = 0

2x(2x + 5) - 3(2x + 5) = 0

(2x + 5) (2x - 3) = 0

2x - 3 = 0 or 2x - 3 = 0

x = 3/2, - 5/2  (As x can't be negative)

x = 3/2

But here we get x = 3/2, - 5/2.