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Question:

Find the roots of the following equations: (i) (x - \frac{1}{x} = 3, x \neq 0) (ii) (\frac{1}{x+4} - \frac{1}{x-7} = \frac{11}{30}, x \neq -4, 7)

Solution:

i)Given equation (x - \frac{1}{x} = 3) (\implies x^2 - 1 = 3x \implies x^2 - 3x - 1 = 0) Here, the middle term (-3x) can't be expressed as a sum of two terms such that their product is equal to the product of extreme terms. So, we use the formula of (x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}) Here, (a=1, b=-3, c=-1; \implies x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-1)}}{2(1)}) (\implies x = \frac{3 \pm \sqrt{9+4}}{2} \implies x = \frac{3 \pm \sqrt{13}}{2}) ii) Similarly, For (\frac{1}{(x+4)} - \frac{1}{(x-7)} = \frac{11}{30}) (\implies \frac{(x-7) - (x+4)}{(x+4)(x-7)} = \frac{11}{30}) (\implies \frac{-11}{(x+4)(x-7)} = \frac{11}{30}) (\implies -30 = x^2 - 3x - 28 \implies x^2 - 3x + 2 = 0) Here, the middle term (-3x) can be expressed as the sum of (-2x) and (-x) such that their product ((-2x) \times (-x) = 2x^2) is equal to the product of extreme terms ((x^2 \times 2 = 2x^2)) (\implies x^2 - 2x - x + 2 = 0 \implies x(x - 2) - 1(x - 2) = 0 \implies (x - 2)(x - 1) = 0 \implies x = 1, 2)