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Solve the following quadratic equations by factorization:\(3(\frac{7x+1}{5x-3})-4(\frac{5x-3}{7x+1})=11;x\neq\frac{3}{5},-\frac{1}{7}\) |
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Answer» In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized. \(3(\frac{7x+1}{5x-3})-4(\frac{5x-3}{7x+1})=11;x\neq\frac{3}{5},-\frac{1}{7}\) ⇒ 3(7x + 1)2 – 4(5x – 3)2 = 11(5x – 3)(7x + 1) ⇒ 3(49x2 + 1 + 14x) – 4(25x2 + 9 – 30x) = 11(35x2 – 3 – 16x) ⇒ 338x2 – 338x = 0 ⇒ x(x – 1) = 0 ⇒ x = 0, 1 |
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