Question

Find roots of the equation using quadratic formula

Answer 1

Dear student
$$\begin{gathered} \frac{1}{\mathrm{x}+1}+\frac{2}{\mathrm{x}+2}=\frac{4}{\mathrm{x}+4} \\ \frac{\mathrm{x}+2+2\left(\mathrm{x}+1\right)}{\left(\mathrm{x}+1\right)\left(\mathrm{x}+2\right)}=\frac{4}{\mathrm{x}+4} \\ \frac{\mathrm{x}+2+2\mathrm{x}+2}{\left(\mathrm{x}+1\right)\left(\mathrm{x}+2\right)}=\frac{4}{\mathrm{x}+4} \\ \frac{3\mathrm{x}+4}{\left(\mathrm{x}+1\right)\left(\mathrm{x}+2\right)}=\frac{4}{\mathrm{x}+4} \\ \left(3\mathrm{x}+4\right)\left(\mathrm{x}+4\right)=4\left({\mathrm{x}}^2+2\mathrm{x}+\mathrm{x}+2\right) \\ 3{\mathrm{x}}^2+12\mathrm{x}+4\mathrm{x}+16=4{\mathrm{x}}^2+12\mathrm{x}+8 \\ 4{\mathrm{x}}^2+12\mathrm{x}+8-3{\mathrm{x}}^2-16\mathrm{x}-16=0 \\ {\mathrm{x}}^2-4\mathrm{x}-8=0 \\ \mathrm{x}=\frac{-\left(-4\right)\pm \sqrt{{\left(-4\right)}^2-4\left(1\right)\left(-8\right)}}{2\left(1\right)}\left[\mathrm{using}\mathrm{quadractic}\mathrm{formula}\mathrm{where}\mathrm{b}=-4,\mathrm{a}=1\mathrm{and}\mathrm{c}=-8\right] \\ =\frac{4\pm \sqrt{16+32}}{2} \\ =\frac{4\pm 4\sqrt{3}}{2} \\ =2\pm 2\sqrt{3} \\ =2\left(1\pm \sqrt{3}\right) \end{gathered}$$
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