Question

Plz tell me this oues

Answer 1

Answer:

hey here is your solution

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Step-by-step explanation:

$so \: here \: given \: equation \: is \\ (1/x) + (1/x + 3) = 40/3 \\ x + 3 + x/x(x + 3) = 40/3 \\ 2x + 3/x {}^{2} + 3x = 40/3 \\ 40( {x}^{2} + 3x) = 3(2x + 3) \\ 40x {}^{2} + 120x = 6x + 9 \\ ie \: \: 40x {}^{2} + 114x - 9 = 0$

$so \: comparing \: this \: quadratic \: equation \: with \: ax {}^{2} + bx + c = 0 \\ we \: get \\ a = 40 \: b = 114 \: c = - 9 \\ now \: using \: \\ Δ = b {}^{2} - 4ac \\ = (114) {}^{2} - (4 \times 40 \times ( - 9)) \\ = 12996 - (160 \times ( - 9)) \\ = 12996 - ( - 1440) \\ = 12996 + 1440 \\ Δ = 14436$

$now \: \: by \: using \: formula \: method \\ we \: get \\ x = - b± \: \sqrt{b {}^{2} - 4ac} / \: 2a \\ = - 114 \: ± \: \sqrt{14336} / \: (2 \times 40) \\ = - 114 \: ± \: 6 \sqrt{401} / \: 80 \\ ie \: \: x = - 114 + 6 \sqrt{401} /80 \: \\ or \: \\ x = - 114 - 6 \sqrt{401} / \: 80 \\ ie \: \: \\ x = - 57 + 3 \sqrt{401} /40 \: \: \\ or \: \\ x = - 57 - 3 \sqrt{401} /40$

Answer 2

$\large\sf{\underline{Solution:-}}$

$\rm\implies \:\dfrac{1}{x}\:+\:\dfrac{1}{x+3}\:=\;\dfrac{40}{3}$

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Taking LCM

$\rm\implies\:\dfrac{(x\: +\: 3) + x}{x(x\:+\:3)}\:=\:\dfrac{40}{3}$

$\rm\implies\:\dfrac{x\: +\: 3\: +\: x}{ {x}^{2} \:+\: 3x}\:=\:\dfrac{40}{3}$

$\rm\implies\:\dfrac{2x\: + \:3 }{ {x}^{2} \:+ \:3x}\:=\:\dfrac{40}{3}$

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Now cross multiplication

$\rm\implies \:(\: 2x \: + \:3\: )\: 3 \: = \: 40 \: ( \: {x}^{2} \: + \: 3x \: )$

$\rm\implies \: 6x \: + \: 9 \: = \: 40 \: {x}^{2} \: + \: 120 \: x \:$

$\rm\implies \: 0\: = \: 40 \: {x}^{2} \: + \: 120 \: x \: - 6 \: x \: - \: 9$

$\rm\implies \: 0\: = \: 40 \: {x}^{2} \: + \: 114 \: x \: - \: 9$

Now since we got a quadratic equation, we will solve it using quadratic formula.

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Quadratic formula is,

$\bf x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

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Here,

• x = roots
• a = coefficient of x²
• b = coefficient of x
• c = constant term

Applying this formula is above quadratic equation, we get . . .

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${\longrightarrow \: \rm x = \dfrac{-114\pm\sqrt{(114)^2-4(40)( - 9)}}{2(40)}}$

${\longrightarrow \: \rm x = \dfrac{-114\pm\sqrt{12996 + 1440}}{80}}$

${\longrightarrow \: \rm x = \dfrac{-114\pm\sqrt{14436}}{80}}$

${\longrightarrow \: \rm x = \dfrac{-114 \: \pm \: 6 \: \sqrt{401}}{80}}$

${\longrightarrow \: \rm x = \dfrac{2(-57 \: \pm \: 3 \: \sqrt{401})}{80}}$

${\longrightarrow \: \rm x = \dfrac{-57 \: \pm \: 3 \: \sqrt{401}}{40}}$

Since this cannot be solved further, this is the required answer !!