Question Number 223823 by fantastic last updated on 06/Aug/25
$$\sqrt{\mathrm{4}{x}+\mathrm{1}}+\sqrt{\mathrm{3}{x}−\mathrm{2}}=\mathrm{1} \\ $$$${x}=? \\ $$
Answered by Rasheed.Sindhi last updated on 06/Aug/25
![(√(4x+1))+(√(3x−2))=1...i (4x+1)−(3x−2)=(√(4x+1)) −(√(3x−2)) (√(4x+1)) −(√(3x−2)) =x+3....ii i+ii: 2(√(4x+1)) =x+4 4(4x+1)=x^2 +8x+16 x^2 −8x+12=0 (x−2)(x−6)=0 x=2, 6 [both don′t satisfy] The equation has no solution.](https://www.tinkutara.com/question/Q223828.png)
$$\sqrt{\mathrm{4}{x}+\mathrm{1}}+\sqrt{\mathrm{3}{x}−\mathrm{2}}=\mathrm{1}…{i} \\ $$$$\left(\mathrm{4}{x}+\mathrm{1}\right)−\left(\mathrm{3}{x}−\mathrm{2}\right)=\sqrt{\mathrm{4}{x}+\mathrm{1}}\:−\sqrt{\mathrm{3}{x}−\mathrm{2}}\: \\ $$$$\sqrt{\mathrm{4}{x}+\mathrm{1}}\:−\sqrt{\mathrm{3}{x}−\mathrm{2}}\:={x}+\mathrm{3}….{ii} \\ $$$${i}+{ii}: \\ $$$$\mathrm{2}\sqrt{\mathrm{4}{x}+\mathrm{1}}\:={x}+\mathrm{4} \\ $$$$\mathrm{4}\left(\mathrm{4}{x}+\mathrm{1}\right)={x}^{\mathrm{2}} +\mathrm{8}{x}+\mathrm{16} \\ $$$${x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{12}=\mathrm{0} \\ $$$$\left({x}−\mathrm{2}\right)\left({x}−\mathrm{6}\right)=\mathrm{0} \\ $$$${x}=\mathrm{2},\:\mathrm{6}\:\left[{both}\:{don}'{t}\:{satisfy}\right] \\ $$$$\:{The}\:{equation}\:{has}\:{no}\:{solution}. \\ $$
Commented by fantastic last updated on 06/Aug/25
$$\vee.\cap¡\subset\in! \\ $$
Answered by mr W last updated on 06/Aug/25
$$\mathrm{3}{x}−\mathrm{2}\geqslant\mathrm{0}\:\Rightarrow{x}\geqslant\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${both}\:\sqrt{\mathrm{4}{x}+\mathrm{1}}\:{and}\:\sqrt{\mathrm{3}{x}−\mathrm{2}}\:{are}\:{strictly} \\ $$$${increasing}. \\ $$$$\left(\sqrt{\mathrm{4}{x}+\mathrm{1}}+\sqrt{\mathrm{3}{x}−\mathrm{2}}\right)_{{min}} \\ $$$$\:\:\:\:=\sqrt{\mathrm{4}×\frac{\mathrm{2}}{\mathrm{3}}+\mathrm{1}}+\sqrt{\mathrm{0}}=\sqrt{\frac{\mathrm{11}}{\mathrm{3}}}>\mathrm{1} \\ $$$$\Rightarrow\sqrt{\mathrm{4}{x}+\mathrm{1}}+\sqrt{\mathrm{3}{x}−\mathrm{2}}=\mathrm{1}\:{has}\:{no}\:{real}\:{root} \\ $$
Commented by fantastic last updated on 06/Aug/25
$${thanks}\:{sir} \\ $$
Answered by MathematicalUser2357 last updated on 11/Aug/25
$$ \\ $$$$ \\ $$