Question

Quadratic equations- Solve in the factorisation method :- x/x+1+x+1/x=34/15 Don't give unnecessary answers and please don't spam.

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}

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Answer 2

Answer:

Given

x/x+1+x+1/x =34/15 , x not equals to 0 , x not equals to -1 .

⇒ x^2+ (x+1)^2/ x (x+1) = 34/15

⇒34x^2+34x^2 =15x^2+15x^2+30x+15

⇒4x^2 +4x−15=0

⇒2x(2x+5)−3(2x+5)=0

⇒2x−3=0 or 2x+5=0

⇒x= 3/2, x=− 5/2