Question

Which of the following are the roots of $3x^2+2x-1=0$ ?
(i) −1
(ii) $\frac{1}{3}$
(iii) $-\frac{1}{2}$

Answer 1

$$\begin{gathered} {\text{The given equation is (3x}}^2+2\mathrm{x}-1=0). \\ \left(\mathrm{i}\right)\mathrm{x}=(-1) \\ \mathrm{L}\mathrm{.H}.S.={\mathrm{x}}^2+2\mathrm{x}-1 \\ =3\times {(-1)}^2+2\times (-1)-1 \\ \mathrm{=}3-2-1 \\ =0 \\ =\mathrm{R}\mathrm{.H}\mathrm{.S}\mathrm{.} \\ \mathrm{Thus},(-1)is\; a\mathrm{root}\mathrm{of}{(3x}^2+2\mathrm{x}-1=0). \\ \left(\mathrm{ii}\right)\mathrm{On\; substituting}\mathrm{x}=\frac{1}{3}\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{equation},\mathrm{we}\mathrm{get}: \\ \mathrm{L}\mathrm{.H}.S.=3{\mathrm{x}}^2+2\mathrm{x}-1 \\ =3\times {\left(\frac{1}{3}\right)}^2+2\times \frac{1}{3}-1 \\ ={\cancel{3}}^1\times \frac{1}{{\sout{9}}_3}+\frac{2}{3}-1 \\ \mathrm{=}\frac{1+2-3}{3} \\ =\frac{0}{3} \\ =0 \\ =\mathrm{R}\mathrm{.H}\mathrm{.S}\mathrm{.} \\ \mathrm{Thus},\left(\frac{1}{3}\right)is\mathrm{a}\mathrm{root}\mathrm{of}{(3x}^2+2\mathrm{x}-1=0). \\ \left(\mathrm{iii}\right)\mathrm{On\; substituting}\mathrm{x}=\left(-\frac{1}{2}\right)\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{equation},\mathrm{we}\mathrm{get}: \\ \mathrm{L}\mathrm{.H}.S.=3{\mathrm{x}}^2+2\mathrm{x}-1 \\ =3\times {\left(-\frac{1}{2}\right)}^2+{\sout{2}}^1\times \left(-\frac{1}{{\sout{2}}_1}\right)-1 \\ =3\times \frac{1}{4}-1-1 \\ \mathrm{=}\frac{3}{4}-2 \\ \mathrm{=}\frac{3-8}{4} \\ =\frac{-5}{4}\ne 0 \\ \mathrm{Thus,\; L}\mathrm{.H}\mathrm{.S}.\cancel{=}\mathrm{R}\mathrm{.H}\mathrm{.S}\mathrm{.} \\ \mathrm{Hence},\left(-\frac{1}{2}\right)is\; a\; not\; solution\; of(3{\mathrm{x}}^2+2\mathrm{x}-1=0). \end{gathered}$$