Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 173051 by mr W last updated on 05/Jul/22

$$solve$$$$(x-2)(x-3)(x-4)(x-6)=5x^2$$

Answered by dragan91 last updated on 06/Jul/22

$$(x^2-8x+12)(x^2-7x+12)-5x^2=0$$$$(x^2-8x+12)((x^2-8x+12)+x)-5(x^2=0$$$${(x^2-8x+12)}^2+2\frac{x}{2}(x^2-8x+12)+\frac{x^2}{4}-\frac{21x^2}{4}=0$$$${(x^2-8x+12+\frac{x}{2})}^2={\left(\frac{x\sqrt{21}}{2}\right)}^2$$$$2x^2-15x+24=\mp x\sqrt{21}$$$$x_1\approx 1,4363$$$$x_2\approx 8,3550$$$$x_{3,4}\approx 2,6044\mp 2,2841i$$$$$$$$$$

Answered by MJS_new last updated on 06/Jul/22

$$\mathrm{we}\mathrm{have}\mathrm{to}\mathrm{solve}\mathrm{the}{4}^{\mathrm{th}}\mathrm{degree}\mathrm{the}\mathrm{usual}\mathrm{way}$$$$\mathrm{I}\mathrm{get}$$$$x_{1,2}=\frac{15+\sqrt{21}\pm \sqrt{6(9+5\sqrt{21})}}{4}$$$$x_{3,4}=\frac{15-\sqrt{21}}{4}\pm \frac{\sqrt{6(-9+5\sqrt{21})}}{4}\mathrm{i}$$

Commented by mr W last updated on 06/Jul/22

$$thankssirs!$$

Commented by dragan91 last updated on 06/Jul/22

$$yes.Ihavesamebutusedaproximatesolutions$$

Commented by mr W last updated on 06/Jul/22

$$yes.youbotharerightsirs.$$

Answered by mr W last updated on 06/Jul/22

$$(x-2)(x-6)(x-3)(x-4)=5x^2$$$$(x^2-8x+12)(x^2-7x+12)=5x^2$$$$(x+\frac{12}{x}-8)(x+\frac{12}{x}-7)=5$$$$letu=x+\frac{12}{x}⩾2\sqrt{12}=4\sqrt{2}or⩽-4\sqrt{2}$$$$(u-8)(u-7)=5$$$$u^2-15u+51=0$$$$u=\frac{15+\sqrt{21}}{2}(\frac{15-\sqrt{21}}{2}<4\sqrt{2}\Rightarrow rejected)$$$$x+\frac{12}{x}=u=\frac{15+\sqrt{21}}{2}$$$$x^2-ux+12=0$$$$x=\frac{1}{2}(u\pm \sqrt{u^2-48})$$$$x=\frac{1}{2}(u\pm \sqrt{15u-99})$$$$x=\frac{1}{2}(\frac{15+\sqrt{21}}{2}\pm \sqrt{15\times \frac{15+\sqrt{21}}{2}-99})$$$$\Rightarrow x=\frac{15+\sqrt{21}\pm \sqrt{6(9+5\sqrt{31})}}{4}✓$$

Commented by Tawa11 last updated on 06/Jul/22

$$\mathrm{Great}\mathrm{sir}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com