Using properties of proportion, solve for x :
3x+9x2−53x−9x2−5=5\dfrac{3x + \sqrt{9x^2 - 5}}{3x - \sqrt{9x^2 - 5}} = 53x−9x2−53x+9x2−5=5
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Applying componendo and dividendo,
⇒3x+9x2−5+3x−9x2−53x+9x2−5−(3x−9x2−5)=5⇒6x29x2−5=5+15−1⇒6x29x2−5=64⇒6x29x2−5=32⇒2x9x2−5=1⇒9x2−5=2x\Rightarrow \dfrac{3x + \sqrt{9x^2 - 5} + 3x - \sqrt{9x^2 - 5}}{3x + \sqrt{9x^2 - 5} - (3x - \sqrt{9x^2 - 5})} = 5 \\[1em] \Rightarrow \dfrac{6x}{2\sqrt{9x^2 - 5}} = \dfrac{5 + 1}{5 - 1} \\[1em] \Rightarrow \dfrac{6x}{2\sqrt{9x^2 - 5}} = \dfrac{6}{4} \\[1em] \Rightarrow \dfrac{6x}{2\sqrt{9x^2 - 5}} = \dfrac{3}{2} \\[1em] \Rightarrow \dfrac{2x}{\sqrt{9x^2 - 5}} = 1 \\[1em] \Rightarrow \sqrt{9x^2 - 5} = 2x \\[1em]⇒3x+9x2−5−(3x−9x2−5)3x+9x2−5+3x−9x2−5=5⇒29x2−56x=5−15+1⇒29x2−56x=46⇒29x2−56x=23⇒9x2−52x=1⇒9x2−5=2x
Squaring both sides of the equation,
⇒9x2−5=4x2⇒9x2−4x2=5⇒5x2=5⇒x2=1⇒x=±1.\Rightarrow 9x^2 - 5 = 4x^2 \\[1em] \Rightarrow 9x^2 - 4x^2 = 5 \\[1em] \Rightarrow 5x^2 = 5 \\[1em] \Rightarrow x^2 = 1 \\[1em] \Rightarrow x = \pm1.⇒9x2−5=4x2⇒9x2−4x2=5⇒5x2=5⇒x2=1⇒x=±1.
Hence, x = ±1.
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If a, b and c are in continued proportion, prove that :
a2+ab+b2b2+bc+c2=ac\dfrac{a^2 + ab + b^2}{b^2 + bc + c^2} = \dfrac{a}{c}b2+bc+c2a2+ab+b2=ca
x+5+x−16x+5−x−16=73.\dfrac{\sqrt{x + 5} + \sqrt{x - 16}}{\sqrt{x + 5} - \sqrt{x - 16}} = \dfrac{7}{3}.x+5−x−16x+5+x−16=37.
If x = a+3b+a−3ba+3b−a−3b\dfrac{\sqrt{a + 3b} + \sqrt{a - 3b}}{\sqrt{a + 3b} - \sqrt{a - 3b}}a+3b−a−3ba+3b+a−3b, prove that :
3bx2 - 2ax + 3b = 0
Using the properties of proportion, solve for x,
given x4+12x2=178\dfrac{x^4 + 1}{2x^2} = \dfrac{17}{8}2x2x4+1=817
