Question

x/x+1 + x+1/x =13/6​

Answer 1

Answer:

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Answer 2

Question:-

➡ Solve the given quadratic equation.

Answer:-

➡ The roots of the given equation are 2 and -3

Solution:-

$\sf \frac{x}{x + 1} + \frac{x + 1}{x} = \frac{13}{6}$

$\sf \implies \frac{(x + 1)^{2} + {x}^{2} }{x(x + 1)} = \frac{13}{6}$

$\sf \implies \frac{ {x}^{2} + 2x + 1 + {x}^{2} }{ {x}^{2} + x } = \frac{13}{6}$

$\sf \implies \frac{ 2{x}^{2} + 2x + 1}{ {x}^{2} + x } = \frac{13}{6}$

On cross multiplying, we get,

$\sf \implies 6( 2{x}^{2} + 2x + 1) = 13({x}^{2} + x )$

$\sf \implies12{x}^{2} + 12x + 6 = 13{x}^{2} + 13x$

Taking everything on left side, we get,

$\sf \implies {x}^{2} + x - 6 =0$

Now, it is in standard form, we will solve it,

$\sf \implies {x}^{2} - 2x + 3x- 6 =0$

$\sf \implies x(x - 2) + 3(x - 2) =0$

$\sf \implies (x + 3)(x - 2) =0$

By zero product rule,

Either x + 3 = 0 or x - 2 = 0

Hence,

$\sf (x + 3)=0 \implies x = - 3$

Again,

$\sf x - 2 = 0 \implies x = 2$

Hence, the roots of the given quadratic equation are 2 and -3.

Verification:-

Let us verify our result.

When x = 2,

$\sf \frac{x}{x + 1} + \frac{x + 1}{x}$

$\sf = \frac{2}{3} + \frac{3}{2}$

$\sf = \frac{4 + 9}{6}$

$\sf = \frac{13}{6}$

Hence, 2 is a root of the equation.

Again,

When x = -3

$\sf \frac{ - 3}{ - 2} + \frac{ - 2}{ - 3}$

$\sf \frac{3}{2} + \frac{2}{3}$

$\sf = \frac{9 + 4}{6}$

$\sf = \frac{13}{6}$Hence, -3 is a root of the equation.

Hence, -3 and 2 are the roots of the given Quadratic Equation (Verified)