Question Number 209633 by Abdullahrussell last updated on 17/Jul/24
Answered by som(math1967) last updated on 17/Jul/24
![(x+1)(3x+2)(6x+5)^2 =6 ⇒(3x^2 +5x+2)(36x^2 +60x+25)=6 ⇒(a+2)(12a+25)=6 [let 3x^2 +5x=a] ⇒12a^2 +49a+50−6=0 ⇒12a^2 +49a+44=0 ⇒12a^2 +16a+33a+44=0 ⇒4a(3a+4)+11(3a+4)=0 ⇒(3a+4)(4a+11)=0 (9x^2 +15x+4)(12x^2 +20x+11)=0 if (9x^2 +15x+4)=0 (3x)^2 +2.3x.(5/2)+((25)/4)=((25)/4)−4 ∴(3x+(5/2))^2 =(9/4)](https://www.tinkutara.com/question/Q209635.png)
$$\left({x}+\mathrm{1}\right)\left(\mathrm{3}{x}+\mathrm{2}\right)\left(\mathrm{6}{x}+\mathrm{5}\right)^{\mathrm{2}} =\mathrm{6} \\ $$$$\Rightarrow\left(\mathrm{3}{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{2}\right)\left(\mathrm{36}{x}^{\mathrm{2}} +\mathrm{60}{x}+\mathrm{25}\right)=\mathrm{6} \\ $$$$\Rightarrow\left({a}+\mathrm{2}\right)\left(\mathrm{12}{a}+\mathrm{25}\right)=\mathrm{6} \\ $$$$\left[{let}\:\mathrm{3}{x}^{\mathrm{2}} +\mathrm{5}{x}={a}\right] \\ $$$$\Rightarrow\mathrm{12}{a}^{\mathrm{2}} +\mathrm{49}{a}+\mathrm{50}−\mathrm{6}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{12}{a}^{\mathrm{2}} +\mathrm{49}{a}+\mathrm{44}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{12}{a}^{\mathrm{2}} +\mathrm{16}{a}+\mathrm{33}{a}+\mathrm{44}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{4}{a}\left(\mathrm{3}{a}+\mathrm{4}\right)+\mathrm{11}\left(\mathrm{3}{a}+\mathrm{4}\right)=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{3}{a}+\mathrm{4}\right)\left(\mathrm{4}{a}+\mathrm{11}\right)=\mathrm{0} \\ $$$$\left(\mathrm{9}{x}^{\mathrm{2}} +\mathrm{15}{x}+\mathrm{4}\right)\left(\mathrm{12}{x}^{\mathrm{2}} +\mathrm{20}{x}+\mathrm{11}\right)=\mathrm{0} \\ $$$${if}\:\left(\mathrm{9}{x}^{\mathrm{2}} +\mathrm{15}{x}+\mathrm{4}\right)=\mathrm{0} \\ $$$$\:\left(\mathrm{3}{x}\right)^{\mathrm{2}} +\mathrm{2}.\mathrm{3}{x}.\frac{\mathrm{5}}{\mathrm{2}}+\frac{\mathrm{25}}{\mathrm{4}}=\frac{\mathrm{25}}{\mathrm{4}}−\mathrm{4} \\ $$$$\therefore\left(\mathrm{3}{x}+\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$ \\ $$
Answered by MM42 last updated on 17/Jul/24
$$\frac{\mathrm{1}}{\mathrm{6}}\left[\left(\mathrm{6}{x}+\mathrm{5}\right)+\mathrm{1}\right)\frac{\mathrm{1}}{\mathrm{2}}\left[\left(\mathrm{6}{x}+\mathrm{5}\right)−\mathrm{1}\right)\left(\mathrm{6}{x}+\mathrm{5}\right)^{\mathrm{2}} =\mathrm{6} \\ $$$$\left({u}+\mathrm{1}\right)\left({u}−\mathrm{1}\right)\left({u}\right)^{\mathrm{2}} =\mathrm{72} \\ $$$$\left({u}^{\mathrm{2}} −\mathrm{1}\right){u}^{\mathrm{2}} −\mathrm{72}= \\ $$$${u}^{\mathrm{4}} −{u}^{\mathrm{2}} −\mathrm{72}=\mathrm{0} \\ $$$$\left({u}^{\mathrm{2}} −\mathrm{9}\right)\left({u}^{\mathrm{2}} +\mathrm{8}\right)=\mathrm{0} \\ $$$${u}=\pm\mathrm{3}\:\:\&\:\:{u}=\pm\mathrm{2}\sqrt{\mathrm{2}}{i} \\ $$$$\left(\mathrm{3}{x}+\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{6}{x}+\mathrm{5}\right)^{\mathrm{2}} =\frac{\mathrm{9}}{\mathrm{4}}\:\:\checkmark\:\&\:\:−\mathrm{2}\:\checkmark \\ $$$$ \\ $$