Solve the following equation by factorisation:
(1+1x+1)(1−1x−1)=78\Big(1 + \dfrac{1}{x + 1}\Big)\Big(1 - \dfrac{1}{x - 1}\Big) = \dfrac{7}{8}(1+x+11)(1−x−11)=87
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Given,
⇒(1+1x+1)(1−1x−1)=78⇒x+1+1x+1×x−1−1x−1=78⇒x+2x+1×x−2x−1=78⇒x2−4x2−1=78⇒8(x2−4)=7(x2−1)⇒8x2−32=7x2−7⇒8x2−7x2=−7+32⇒x2=25⇒x=25⇒x=±5.\Rightarrow \Big(1 + \dfrac{1}{x + 1}\Big)\Big(1 - \dfrac{1}{x - 1}\Big) = \dfrac{7}{8} \\[1em] \Rightarrow \dfrac{x + 1 + 1}{x + 1} \times \dfrac{x - 1 - 1}{x - 1} = \dfrac{7}{8} \\[1em] \Rightarrow \dfrac{x + 2}{x + 1} \times \dfrac{x - 2}{x - 1} = \dfrac{7}{8} \\[1em] \Rightarrow \dfrac{x^2 - 4}{x^2 - 1} = \dfrac{7}{8} \\[1em] \Rightarrow 8(x^2 - 4) = 7(x^2 - 1) \\[1em] \Rightarrow 8x^2 - 32 = 7x^2 - 7 \\[1em] \Rightarrow 8x^2 - 7x^2 = - 7 + 32 \\[1em] \Rightarrow x^2 = 25 \\[1em] \Rightarrow x = \sqrt{25} \\[1em] \Rightarrow x = \pm 5.⇒(1+x+11)(1−x−11)=87⇒x+1x+1+1×x−1x−1−1=87⇒x+1x+2×x−1x−2=87⇒x2−1x2−4=87⇒8(x2−4)=7(x2−1)⇒8x2−32=7x2−7⇒8x2−7x2=−7+32⇒x2=25⇒x=25⇒x=±5.
Hence, value of x = -5 or +5.
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4x+2−1x+3=42x+1\dfrac{4}{x + 2} - \dfrac{1}{x + 3} = \dfrac{4}{2x + 1}x+24−x+31=2x+14
5x−2−3x+6=4x\dfrac{5}{x - 2} - \dfrac{3}{x + 6} = \dfrac{4}{x}x−25−x+63=x4
Find the quadratic equation, whose solution set is :
(i) {3, 5}
(ii) {-2, 3}
Solve : x3+36−x=2(6+x)15\dfrac{x}{3} + \dfrac{3}{6 - x} = \dfrac{2(6 + x)}{15}3x+6−x3=152(6+x); (x ≠ 6)
