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 210996 by RojaTaniya last updated on 25/Aug/24

Answered by Frix last updated on 26/Aug/24

$$\mathrm{Only}\mathrm{true}\mathrm{for}\mathrm{the}\mathrm{real}\mathrm{solution}>0:$$$$x^3+2x^2+5x-1=0$$$$x^3+6x^2+9x=4x^2+4x+1$$$$x{(x+3)}^2={(2x+1)}^2$$$$x^{\frac{1}{2}}(x+3)=2x+1$$$$x^{\frac{3}{2}}+3x^{\frac{1}{2}}=2x+1$$$$x=1-3x^{\frac{1}{2}}+3x-x^{\frac{3}{2}}$$$$x={(1-x^{\frac{1}{2}})}^3$$$$x^{\frac{1}{3}}=1-x^{\frac{1}{2}}$$$$x^{\frac{1}{2}}+x^{\frac{1}{3}}=1$$

Commented by Rasheed.Sindhi last updated on 26/Aug/24

$$\vee .\cap \mathbf{i}\subset \in !$$

Commented by Frix last updated on 26/Aug/24

��

Answered by Frix last updated on 26/Aug/24

$$\mathrm{Another}\mathrm{method},\mathrm{not}\mathrm{recommended},\mathrm{thus}$$$$\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{write}\mathrm{it}\mathrm{out}...$$$$(x^3+2x^2+5x-1):(x^{\frac{1}{2}}+x^{\frac{1}{3}}-y)$$$$(t^{18}+2t^{12}+5t^6-1):(t^3+t^2-y)=$$$$=\underset{j=0}{\sum ^{15}}\left(c_j{t}^j\right)-\frac{y-1}{t^3+t^2-y}\underset{k=0}{\sum ^2}\left(d_k{t}^k\right)$$$$\Rightarrow y=1$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com