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    Which of the options below give the solution to the equation ​1x+3+1x+5=16\frac{1}{x+3} + \frac{1}{x+5} = \frac{1}{6}x+31​+x+51​=61​ ?​​
    Question

    Which of the options below give the solution to the equation ​1x+3+1x+5=16\frac{1}{x+3} + \frac{1}{x+5} = \frac{1}{6} ?​​

    A.

    ​x=2±37x = 2 \pm \sqrt{37}​​

    B.

    ​x=2±39x = 2 \pm \sqrt{39}​​

    C.

    ​x=3±37x = 3 \pm \sqrt{37}​​

    D.

    ​x=4±37 x = 4 \pm \sqrt{37}​​

    Correct option is A

    Given : 

    1x+3+1x+5=16\frac{1}{x+3} + \frac{1}{x+5} = \frac{1}{6}

    Formula Used: 

    For quadratic equation  ax2+bx+c=0ax^2+bx+c=0​

    quadratic formula =  x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}​​

    Solution : 

    1x+3+1x+5\frac{1}{x+3} + \frac{1}{x+5}

    =(x+5)+(x+3)(x+3)(x+5)=\frac{(x+5) + (x+3)}{(x+3)(x+5)}

    =2x+8(x+3)(x+5)=\frac{2x+8}{(x+3)(x+5)}

    ​​​​=2x+8(x+3)(x+5)=16=\frac{2x+8}{(x+3)(x+5)} = \frac{1}{6}

    = 6 (2x+8) = (x+3) (x+5)

    ​=12x+48=x2+8x+15=12x + 48 = x^2 + 8x + 15

    =x2−4x−33=0=x^2 - 4x - 33 = 0

    ​x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

    ​x=−(−4)±(−4)2−4(1)(−33)2(1)x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-33)}}{2(1)}

    x=4±16+1322x = \frac{4 \pm \sqrt{16 + 132}}{2}

    x=4±1482x = \frac{4 \pm \sqrt{148}}{2}

    x=4±2372x = \frac{4 \pm 2\sqrt{37}}{2}

    x=2±37x = 2 \pm \sqrt{37}​​​​​​​​

    ​​

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