Question: Solve: $\dfrac{\sqrt{x+4} + \sqrt{2} }{ \sqrt{x +4} - \sqrt{2}} = 2$ Solution: Given, $\dfrac{\sqrt{x+4} + \sqrt{2} }{ \sqrt{x +4} - \sqrt{2}} = 2$ $or, \sqrt{x+4} + \sqrt{2}  = 2(  \sqrt{x +4} - \sqrt{2})$ $or, \sqrt{x+4} + \sqrt{2} = 2 \sqrt{x +4} - 2\ sqrt{2})$ $or, 2\sqrt{2} + \sqrt{2} = 2\sqrt{x +4} - \sqrt{x +4}$ $o…

Question: Solve: $\dfrac{√x +√5}{√x -√5} + \dfrac{√x -√5}{√x +√5} = 4$ Solution: Given, $\dfrac{√x +√5}{√x -√5} + \dfrac{√x -√5}{√x +√5} = 4$ $or, \dfrac{(√x +√5)(√x +√5) + (√x -√5)(√x -√5)}{(√x)² - (√5)²} = 4$ $or, \dfrac{(√x)² + 2×√x×√5 + (√5)² + (√x)² -2×√x*√5 +(√5)²}{x -5} = 4$ $or, x + 2\sqrt{5x} + 5 + x -2\sqrt{5x} + 5 = 4(x…

Question: Solve: $\dfrac{\sqrt{x +2} - \sqrt{x -2}}{\sqrt{x +2} + \sqrt{x -2}} = \dfrac{1}{2}$ Solution: Given, $\dfrac{\sqrt{x +2} - \sqrt{x -2}}{\sqrt{x +2} +\sqrt{x -2}} = \dfrac{1}{2}$ $or, 2(\sqrt{x +2} - \sqrt{x -2}) = 1 ( \sqrt{x +2} +\sqrt{x -2})$ $or, 2\sqrt{x +2} - 2 \sqrt{x -2} = \sqrt{x +2} + \sqrt{x -2}$ $or, 2\sqrt{x…

Question: Solve: $\dfrac{3x-4}{2+ \sqrt{3x}} - \dfrac{\sqrt{3x} -2 }{2} = 2$ Solution: Given, $\dfrac{3x-4}{2+ \sqrt{3x}} - \dfrac{\sqrt{3x} -2 }{2} = 2$ $or, \dfrac{(\sqrt{3x})^2 - 2^2}{\sqrt{3x} + 2}  - \dfrac{\sqrt{3x} -2 }{2} = 2$ $or, \dfrac{(\sqrt{3x} +2)(\sqrt{3x} - 2)}{\sqrt{3x} + 2}  - \dfrac{\sqrt{3x} -2 }{2} = 2$ $or, (\s…

Question: Solve $\dfrac{y-25}{5+\sqrt{y}} = 4 + \dfrac{\sqrt{y} -5 }{5}$ Solution: Given, $\dfrac{y-25}{5+\sqrt{y}} = 4 + \dfrac{\sqrt{y} -5 }{5}$ $or, \dfrac{(\sqrt{y})^2 - 5^2}{\sqrt{y} + 5} = 4 + \dfrac{\sqrt{y} -5 }{5}$ $or, \dfrac{(\sqrt{y} +5)(\sqrt{x} -5)}{\sqrt{y} + 5} = 4 + \dfrac{\sqrt{y} -5 }{5}$ $or, \sqrt{y} - 5 = 4+ \d…

Question: Solve: $\dfrac{x-1}{\sqrt{x} +1 } = 4 + \dfrac{\sqrt{x} -1}{2}$ Solution: Given, $\dfrac{x-1}{\sqrt{x} +1 } = 4 + \dfrac{\sqrt{x} -1}{2}$ $or, \dfrac{(\sqrt{x})^2 - (1)^2}{\sqrt{x} +1 } =  4+ \dfrac{\sqrt{x} -1}{2}$ $or, \dfrac{(\sqrt{x} +1 )(\sqrt{x} -1)}{(\sqrt{x) +1)} = 4 + \dfrac{\sqrt{x} -1}{2}$ $or, \sqrt{x} - 1 = 4 …

Question: Solve: $\sqrt{x} + \sqrt{x+13} = \dfrac{91}{\sqrt{x+13}}$ Solution: Given,  $\sqrt{x} + \sqrt{x+13} = \dfrac{91}{\sqrt{x+13}}$ $or, \sqrt{x}  = \dfrac{91}{\sqrt{x+13}} - \sqrt{x+13}$ $or, \sqrt{x} = \dfrac{91 - (\sqrt{x+13})(\sqrt{x+13})}{\sqrt{x+13}}$ $or, \sqrt{x} (\sqrt{x + 13} = 91 - \sqrt{(x+13)^2}$ $or, \sqrt{x(x+13)…

Question: Solve: $2\sqrt{x} - \sqrt{4x -3} = \dfrac{1}{\sqrt{4x-3}}$ Solution: Given, $2\sqrt{x} - \sqrt{4x -3} = \dfrac{1}{\sqrt{4x-3}}$ $or, 2\sqrt{x} = \dfrac{1}{\sqrt{4x-3}} + \sqrt{4x -3}$ $or, 2\sqrt{x} = \dfrac{1 + (\sqrt{4x-3})(\sqrt{4x-3})}{\sqrt{4x-3}}$ $or, 2\sqrt{x}(\sqrt{4x-3}) = 1 + \sqrt{(4x-3)(4x-3)}$ $or, 2\sqrt{x(4…

Question: Solve: $\sqrt{x} + \sqrt{5+x} = \dfrac{15}{\sqrt{5+x}}$ Solution: Given, $\sqrt{x} + \sqrt{5+x} = \dfrac{15}{\sqrt{5+x}}$ $or, \sqrt{x} = \dfrac{15}{\sqrt{5+x} }- \sqrt{5+x}$ $or, \sqrt{x} = \dfrac{15 - (\sqrt{5+x})(\sqrt{5+x})}{\sqrt{5+x}}$ $or, \sqrt{x} = \dfrac{15 - \sqrt{(5+x)(5+x)}}{\sqrt{5+x}}$ $or, \sqrt{x} (\sqrt{5…

Question: Solve: $\sqrt{4x -3} + \sqrt{2x +3} = 6$ Solution: Given, $\sqrt{4x -3} + \sqrt{2x +3} = 6$ $or, \sqrt{4x -3} = 6- \sqrt{2x +3}$ Squaring both sides $or, ( \sqrt{4x -3})² = (6- \sqrt{2x +3})²$ $or, 4x -3 = 6² - 2×6×\sqrt{2x +3} + (\sqrt{2x +3})²$ $or, 4x -3 = 36 - 12\sqrt{2x +3} +2x +3$ $or, 4x -3 = 39 + 2x -12\sqrt{2x +3}…

Question: Solve: $\sqrt{3x +1} - \sqrt{x -1} = 2$ Solution: Given, $\sqrt{3x +1} - \sqrt{x -1} = 2$ $or, \sqrt{3x +1} = 2 + \sqrt{x -1}$ [ Squaring both sides ] $or, (\sqrt{3x +1})² = (2+\sqrt{x -1})²$ $or, 3x +1 = 2² + 2×2×\sqrt{x -1} + (\sqrt{x-1})²$ $or, 3x +1 = 4 + 4\sqrt{x -1} + x -1$ $or, 3x -x = 4-1 -1 + 4\sqrt{x -1}$ $or, 2x…

Question: Solve: $2x +1 = \sqrt{4x²+3x +6}$ Solution: Given, $2x +1 = \sqrt{4x²+3x +6}$ [ squaring both sides ] $or, (2x +1)² = (\sqrt{4x²+3x+6})²$ $or, (2x)² + 2×2x×1 +1² = 4x²+3x+6$ $or, 4x² +4x +1 = 4x² + 3x +6$ $or, 4x²-4x²+4x -3x = 6-1$ $\therefore x = 5$

Question: Solve: $\sqrt{3x +4} + x = 12$ Solution: Given, $\sqrt{3x +4} + x = 12$ $or, \sqrt{3x +4} = 12-x$ [ Squaring both sides ] $or, (\sqrt{3x +4})^2 = (12-x)^2$ $or, 3x +4 = 12^2 - 2×12×x + x^2 $ $or, 3x +4 = 144 - 24x + x^2$ $or, x^2 - 24x -3x +144-4 = 0$ $or, x^2 - 27x +140 = 0$ $or, x^2 - (20+7)x +140 = 0$ $or, x^2 - 20x -7x…

Question: Solve: $\sqrt{2x +7} = x +2$ Solution: Given, $\sqrt{2x +7} = x +2$ [ Squaring both sides ] $or, (\sqrt{2x +7})^2 = (x +2)^2$ $or, 2x +7 = x^2 + 2×x×2 + 2^2$ $or, 2x +7 = x^2 +4x + 4$ $or, x^2 + 4x -2x +4 -7 = 0$ $or, x^2 +2x -3 = 0$ $or, x^2 +(3-1)x +3= 0$ $or, x^2 + 3x -x +3 = 0$ $or, x(x +3) -1(x +3) = 0$ $or, (x -1)(x …

Question: Solve: $\dfrac{\sqrt{x} + \sqrt{7}}{\sqrt{x} - \sqrt{7}} = 3$ Solution: Given, $\dfrac{\sqrt{x} + \sqrt{7}}{\sqrt{x} - \sqrt{7}} = 3$ $or, \sqrt{x} - \sqrt{7} = 3(\sqrt{x} - \sqrt{7})$ $or, \sqrt{x} - \sqrt{7}= 3\sqrt{x} - 3\sqrt{7}$ $or, 3\sqrt{7} + \sqrt{7} = 3\sqrt{x} - \sqrt{x}$ $or, 4\sqrt{7} = 2\sqrt{x}$ $or, 2\sqrt{…

Question: Solve: $\dfrac{√x -4}{√x} = \dfrac{3}{7}$ Solution: Given, $\dfrac{√x -4}{√x} = \dfrac{3}{7}$ $or, 7(√x -4) = 3√x$ $or, 7√x - 28 = 3√x$ $or, 7√x - 3√x = 28$ $or, 4√x = 28$ $or, √x = \dfrac{28}{4}$ $or, √x = 7$ [ Squaring both sides ] $or, (√x)^2 = 7^2$ $\therefore x = 49$ = Answer

Question: Solve: $\dfrac{5y -4}{\sqrt{5y} - 2} = 2 - \dfrac{\sqrt{5y} -3}{2}$ Solution: Given, $\dfrac{5y -4}{\sqrt{5y} - 2} = 2 - \dfrac{\sqrt{5y} -3}{2}$ $or, \dfrac{(\sqrt{5y})^2 - 2^2}{\sqrt{5y }-2} = \dfrac{4 - (\sqrt{5y} -3)}{2}$ $or, \dfrac{(\sqrt{5y} +2)(\sqrt{5y} -2)}{\sqrt{5y}-2} = \dfrac{4 - \sqrt{5y} +3}{2}$ $or, \sqrt…

Question: Solve: $\dfrac{x -25}{√x -5} = 9$ Solution: Given, $\dfrac{x -25}{√x -5} = 9$ $or, \dfrac{(√x)^2 - (5)^2}{√x -5} = 9$ $or, \dfrac{(√x +5)(√x -5)}{(√x -5)} = 9$ $or, √x +5 = 9$ $or, √x = 9-5$ $or, √x = 4$ [ Squaring both sides ] $or, (√x)^2 = 4^2$ $\therefore x = 16$ = Answer

Question: Solve: $\dfrac{x -1}{√x +1} = 1$ Solution: Given, $\dfrac{x -1}{√x +1} = 1$ $or, \dfrac{(√x)^2 - (1)^2}{√x +1} = 1$ $or, \dfrac{(√x +1)(√x -1)}{√x +1} = 1$ $or, √x -1 = 1$ $or, √x = 1 +1$ $or, √x = 2$ [ Squaring both sides ] $or, (√x)^2 = 2^2$ $\therefore x = 4$ = Answer

Question: Solve: $\dfrac{x -9}{\sqrt{x}+3} = 1$ Solution: Given, $\dfrac{x -9}{\sqrt{x}+3} = 1$ $or, \dfrac{(\sqrt{x})^2 -(3)^2}{\sqrt{x}+3} = 1$ $or, \dfrac{(\sqrt{x} + 3)(\sqrt{x} - 3) }{\sqrt{x} + 3} = 1$ $or, \sqrt{x} - 3 = 1$ $or, \sqrt{x} = 3 +1$ $or, \sqrt{x} = 4$ [ Squaring both sides ] $or, (\sqrt{x})^2 = 4^2$ $\therefore x…