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Q3. Find the roots of the following equations:
(i) \({{x} - {{1}\over {x}}} = 3, x \neq 0\)
(ii) \({{1} \over {x + 4}} - {{1} \over {x - 7}} = {{11} \over {30}}, x \neq -4,7\)
Answer :

(i)\({{x} - {{1}\over {x}}} = 3 \) where x is not equal to 0
=>\({{x^2 - 1} \over {x}} = 3\)
=>\(x^2 - 1 = 3x\)
=>\(x^2 -3x - 1 = 0\)
The general form of equation is \(ax^2 + bx + c = 0\)
Quadratic formula = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)
=>\(x = {{3 ± \sqrt{(3)^2 - 4(1)(-1)}} \over {(2)(1)}}\)
=>\(x = {{3 ± \sqrt{13}} \over {2}}\)
=>\(x = {{3 + \sqrt{13}} \over {2}} , {{3 - \sqrt{13}} \over {2}}\)

(ii)\({{1} \over {x + 4}} - {{1} \over {x - 7}} = {{11} \over {30}}, x ? -4,7\)
=>\({{(x - 7) - (x + 4)} \over {(x - 4)(x - 7)}} = {{11} \over {30}}\)
=>\({{-11} \over {(x - 4)(x - 7)}} = {{11} \over {30}}\)
=>\(-30 = x^2 - 7x + 4x -28\)
=>\(x^2 -3x + 2 = 0\)
The general form of equation is \(ax^2 + bx + c = 0\)
Quadratic formula = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)
=>\(x = {{3 ± \sqrt{(3)^2 - 4(1)(2)}} \over {(2)(1)}}\)
=>\(x = {{3 ± \sqrt{1}} \over {2}}\)
=>\(x = {{3 + \sqrt{1}} \over {2}} , {{3 - \sqrt{1}} \over {2}}\)
=>\(x = 2,1\)