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 164269 by Zaynal last updated on 15/Jan/22

                                Solve the inequality;    (x+3)[(x−1)^2 −x(x+(3/4))]+6+((7x^2 +29x)/4)>0  {Z.A}

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{Solve}}\:\boldsymbol{{the}}\:\boldsymbol{{inequality}}; \\ $$ $$\:\:\left(\boldsymbol{{x}}+\mathrm{3}\right)\left[\left(\boldsymbol{{x}}−\mathrm{1}\right)^{\mathrm{2}} −\boldsymbol{{x}}\left(\boldsymbol{{x}}+\frac{\mathrm{3}}{\mathrm{4}}\right)\right]+\mathrm{6}+\frac{\mathrm{7}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{29}\boldsymbol{{x}}}{\mathrm{4}}>\mathrm{0} \\ $$ $$\left\{\boldsymbol{{Z}}.\boldsymbol{{A}}\right\} \\ $$

Answered by manxsolar last updated on 15/Jan/22

primer termino raiz x=−3  segundl termino  24+7x^2 +29x=(7x+8)(x+3)/4  factorizando     (x+3)[x^2 −2x+1−x^2 −((3x)/4)+((7x)/4)+(8/4)]⟩0  (x+3)[−2x+1+x+2]⟩0  (x+3)[−x+3]⟩0  por −1  (x+3)(x−3)⟨0  puntoz criticos  x=−3  x..=3  −−−−−−−(−3)−−−−−−−(+3)−−−−−−−              +                                       −                                          +  me piden f(x)⟨0  entonces            x∈<−3;3>  sean x1,x2,x3,........xn    raices del polinomio=(x−x1)(x−x2((x−x3).....                           xn..........x3.......x2.........x1.........         f(x)                                   +                 −              +      condiciones uso  termino x  positivo  x1,x2,x3,     ordenadoz eje carteziano

$${primer}\:{termino}\:{raiz}\:{x}=−\mathrm{3} \\ $$ $${segundl}\:{termino}\:\:\mathrm{24}+\mathrm{7}{x}^{\mathrm{2}} +\mathrm{29}{x}=\left(\mathrm{7}{x}+\mathrm{8}\right)\left({x}+\mathrm{3}\right)/\mathrm{4} \\ $$ $${factorizando} \\ $$ $$\:\:\:\left({x}+\mathrm{3}\right)\left[{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}−{x}^{\mathrm{2}} −\frac{\mathrm{3}{x}}{\mathrm{4}}+\frac{\mathrm{7}{x}}{\mathrm{4}}+\frac{\mathrm{8}}{\mathrm{4}}\right]\rangle\mathrm{0} \\ $$ $$\left({x}+\mathrm{3}\right)\left[−\mathrm{2}{x}+\mathrm{1}+{x}+\mathrm{2}\right]\rangle\mathrm{0} \\ $$ $$\left({x}+\mathrm{3}\right)\left[−{x}+\mathrm{3}\right]\rangle\mathrm{0} \\ $$ $${por}\:−\mathrm{1} \\ $$ $$\left({x}+\mathrm{3}\right)\left({x}−\mathrm{3}\right)\langle\mathrm{0} \\ $$ $${puntoz}\:{criticos} \\ $$ $${x}=−\mathrm{3} \\ $$ $${x}..=\mathrm{3} \\ $$ $$−−−−−−−\left(−\mathrm{3}\right)−−−−−−−\left(+\mathrm{3}\right)−−−−−−− \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:+\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+ \\ $$ $${me}\:{piden}\:{f}\left({x}\right)\langle\mathrm{0} \\ $$ $${entonces}\:\:\:\:\:\:\:\:\:\:\:\:{x}\in<−\mathrm{3};\mathrm{3}> \\ $$ $${sean}\:{x}\mathrm{1},{x}\mathrm{2},{x}\mathrm{3},........{xn}\:\:\:\:{raices}\:{del}\:{polinomio}=\left({x}−{x}\mathrm{1}\right)\left({x}−{x}\mathrm{2}\left(\left({x}−{x}\mathrm{3}\right).....\right.\right. \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{xn}..........{x}\mathrm{3}.......{x}\mathrm{2}.........{x}\mathrm{1}......... \\ $$ $$\:\:\:\:\:\:\:{f}\left({x}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\:\:\:\:\:\:\:\:\:\:\:\:\:\:+ \\ $$ $$ \\ $$ $$ \\ $$ $${condiciones}\:{uso} \\ $$ $${termino}\:{x}\:\:{positivo} \\ $$ $${x}\mathrm{1},{x}\mathrm{2},{x}\mathrm{3},\:\:\:\:\:{ordenadoz}\:{eje}\:{carteziano} \\ $$ $$ \\ $$ $$ \\ $$ $$ \\ $$ $$ \\ $$ $$ \\ $$ $$ \\ $$ $$ \\ $$ $$ \\ $$ $$ \\ $$ $$ \\ $$ $$ \\ $$ $$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com