Question

Solve the following equation by using the general formula , if the equation has a solution R :
i)$\frac{1}{x}-\frac{1}{x-3}=3,x\ne 0,2$
ii)$x+\frac{1}{x}=3,x\ne 0$iii)$3x^2+2\sqrt{5x}-5=0$
iv)$\sqrt{3}{x}^2-2x+\sqrt{3}=0$
v)$\frac{1}{x+1}+\frac{2}{x+2}=\frac{4}{x+4};x\ne 1,-2,-4$
vi)$\sqrt{2}{x}^2+7x+5\sqrt{2=0}$

Answer 1

i) $\frac{1}{x}-\frac{1}{x-3}$

$\Rightarrow x-3-x=3x^2-9x$

$\Rightarrow 3x^2-9x+3=0$

$\Rightarrow x=\frac{9\pm \sqrt{81-36}}{2\left(3\right)}$

$\Rightarrow x=\frac{9+\sqrt{4}5}{6},\frac{4-\sqrt{4}5}{6}$

ii) $x^2-3x+1=0$

$x=\frac{3\pm \sqrt{9-4}}{2}$

$x=\frac{3+\sqrt{5}}{2},\frac{3-\sqrt{5}}{2}$

iii) $3x^2+2\sqrt{5}x-5=0$

$x=\frac{2\sqrt{5}\pm \sqrt{20+60}}{2\left(3\right)}$

$x=\sqrt{5},\frac{-5\sqrt{5}}{3}$

iv)$\sqrt{3}{x}^2+2\sqrt{5}x-5=0$

$D\equiv 4-3\left(4\right)<0$ No $x\notin R$

v) $(x+2)(x+4)+2(x+1)(x+4)=4(x+1)(x+2)$

$\Rightarrow x^2-4x-8=0$

$x=\frac{4\pm \sqrt{16+32}}{2}$

$x=2+8\sqrt{3},2-8\sqrt{3}$

vi) $\sqrt{2}{x}^2+7x+5\sqrt{2}=0$

$x=\frac{-7\pm \sqrt{49-40}}{2\sqrt{2}}$

$x=-\sqrt{2},\frac{-5}{\sqrt{2}}$