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 77035 by necxxx last updated on 02/Jan/20

Solve for x in: (i) (2(x+3)−3(x−2))(2x−1)≥0 (ii)(x−1)(2x+3)(x+1)(x+3)≤1

$${Solve}\:{for}\:{x}\:{in}: \\ $$$$\left({i}\right)\:\left(\mathrm{2}\left({x}+\mathrm{3}\right)−\mathrm{3}\left({x}−\mathrm{2}\right)\right)\left(\mathrm{2}{x}−\mathrm{1}\right)\geqslant\mathrm{0} \\ $$$$\left({ii}\right)\left({x}−\mathrm{1}\right)\left(\mathrm{2}{x}+\mathrm{3}\right)\left({x}+\mathrm{1}\right)\left({x}+\mathrm{3}\right)\leqslant\mathrm{1} \\ $$

Answered by MJS last updated on 03/Jan/20

(i) −2x^2 +25x−12≥0 −2(x−12)(x−(1/2))≥0 (x−12)(x−(1/2))≤0 ⇒ (1/2)≤x≤12 (ii) 2x^4 +9x^3 +7x^2 +9x−10≤0 zeros at x≈−3.03943∧x≈1.02431 ⇒ ≈−3.03943≤x≤≈1.02431

$$\left({i}\right) \\ $$$$−\mathrm{2}{x}^{\mathrm{2}} +\mathrm{25}{x}−\mathrm{12}\geqslant\mathrm{0} \\ $$$$−\mathrm{2}\left({x}−\mathrm{12}\right)\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)\geqslant\mathrm{0} \\ $$$$\left({x}−\mathrm{12}\right)\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)\leqslant\mathrm{0}\:\Rightarrow\:\frac{\mathrm{1}}{\mathrm{2}}\leqslant{x}\leqslant\mathrm{12} \\ $$$$\left({ii}\right) \\ $$$$\mathrm{2}{x}^{\mathrm{4}} +\mathrm{9}{x}^{\mathrm{3}} +\mathrm{7}{x}^{\mathrm{2}} +\mathrm{9}{x}−\mathrm{10}\leqslant\mathrm{0} \\ $$$$\mathrm{zeros}\:\mathrm{at}\:{x}\approx−\mathrm{3}.\mathrm{03943}\wedge{x}\approx\mathrm{1}.\mathrm{02431} \\ $$$$\Rightarrow\:\approx−\mathrm{3}.\mathrm{03943}\leqslant{x}\leqslant\approx\mathrm{1}.\mathrm{02431} \\ $$

Commented by necxxx last updated on 03/Jan/20

Thank you so much MJS .However one of the major places I had issue in was in solving question (ii). Is there a way to solve such quartic equations? Please help me out with that. Thank you so much sir.

$${Thank}\:{you}\:{so}\:{much}\:{MJS}\:.{However}\:{one} \\ $$$${of}\:{the}\:{major}\:{places}\:{I}\:{had}\:{issue}\:{in}\:{was} \\ $$$${in}\:{solving}\:{question}\:\left({ii}\right).\:{Is}\:{there}\:{a}\:{way}\:{to}\: \\ $$$${solve}\:{such}\:{quartic}\:{equations}?\:{Please}\:{help} \\ $$$${me}\:{out}\:{with}\:{that}.\:{Thank}\:{you}\:{so}\:{much}\:{sir}. \\ $$

Commented by MJS last updated on 03/Jan/20

in fact we can only try to find exact solutions (1) (x−a)(x−b)(x−c)(x−d)=x^4 +...+abcd ⇒ try all factors of ±the constant (2) try to find 2 square factors x^4 +ax^3 +bx^2 +cx+d=0; first let x=z−(a/4) leading to z^4 +pz^2 +qz+r=0 now try to find (z^2 −αz−β)(z^2 +αz−γ)=0 by solving the system for α, β, γ { ((p=−(α^2 +β+γ))),((q=α(γ−β))),((r=βγ)) :} starting with (i) γ=... and (ii) β=... this leads to (iii) a polynome of degree 3 for (α^2 ) which sometimes leads to a “nice” solution if not, this doesn′t help (3) some pokynomes of degree 4 can be solved using one of the following “models”. you have to expand and match the constants which leads to systems of 4 equations for α, β, γ, δ. again leading to polynomes of degree 3 for (α^2 ) or (γ^2 ) which might have “nice” solutions (x−α−(√β))(x−α+(√β))(x−γ−(√δ))(x−γ+(√δ))=0 (x−α−(√β)−(√γ)−(√δ))(x−α−(√β)+(√γ)+(√δ))(x−α+(√β)−(√γ)+(√δ))(x−α+(√β)+(√γ)−(√δ))=0 (x+α+(√β)+(√γ)+(√δ))(x+α+(√β)−(√γ)−(√δ))(x+α−(√β)+(√γ)−(√δ))(x+α−(√β)+>−(√γ)+(√δ))=0 if all these don′t work you can only approximate because Ferrari′s solution is always possible to find but never possible to handle

$$\mathrm{in}\:\mathrm{fact}\:\mathrm{we}\:\mathrm{can}\:\mathrm{only}\:\mathrm{try}\:\mathrm{to}\:\mathrm{find}\:\mathrm{exact}\:\mathrm{solutions} \\ $$$$\left(\mathrm{1}\right)\:\left({x}−{a}\right)\left({x}−{b}\right)\left({x}−{c}\right)\left({x}−{d}\right)={x}^{\mathrm{4}} +...+{abcd} \\ $$$$\:\:\:\:\:\:\:\Rightarrow\:\mathrm{try}\:\mathrm{all}\:\mathrm{factors}\:\mathrm{of}\:\pm\mathrm{the}\:\mathrm{constant} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{try}\:\mathrm{to}\:\mathrm{find}\:\mathrm{2}\:\mathrm{square}\:\mathrm{factors} \\ $$$$\:\:\:\:\:\:\:{x}^{\mathrm{4}} +{ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d}=\mathrm{0};\:\mathrm{first}\:\mathrm{let}\:{x}={z}−\frac{{a}}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\mathrm{leading}\:\mathrm{to}\:{z}^{\mathrm{4}} +{pz}^{\mathrm{2}} +{qz}+{r}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\mathrm{now}\:\mathrm{try}\:\mathrm{to}\:\mathrm{find}\:\left({z}^{\mathrm{2}} −\alpha{z}−\beta\right)\left({z}^{\mathrm{2}} +\alpha{z}−\gamma\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\mathrm{by}\:\mathrm{solving}\:\mathrm{the}\:\mathrm{system}\:\mathrm{for}\:\alpha,\:\beta,\:\gamma \\ $$$$\:\:\:\:\:\:\:\begin{cases}{{p}=−\left(\alpha^{\mathrm{2}} +\beta+\gamma\right)}\\{{q}=\alpha\left(\gamma−\beta\right)}\\{{r}=\beta\gamma}\end{cases} \\ $$$$\:\:\:\:\:\:\:\mathrm{starting}\:\mathrm{with}\:\left(\mathrm{i}\right)\:\gamma=...\:\mathrm{and}\:\left(\mathrm{ii}\right)\:\beta=...\:\mathrm{this} \\ $$$$\:\:\:\:\:\:\:\mathrm{leads}\:\mathrm{to}\:\left(\mathrm{iii}\right)\:\mathrm{a}\:\mathrm{polynome}\:\mathrm{of}\:\mathrm{degree}\:\mathrm{3}\:\mathrm{for}\:\left(\alpha^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:\:\mathrm{which}\:\mathrm{sometimes}\:\mathrm{leads}\:\mathrm{to}\:\mathrm{a}\:``\mathrm{nice}''\:\mathrm{solution} \\ $$$$\:\:\:\:\:\:\:\mathrm{if}\:\mathrm{not},\:\mathrm{this}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{help} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{some}\:\mathrm{pokynomes}\:\mathrm{of}\:\mathrm{degree}\:\mathrm{4}\:\mathrm{can}\:\mathrm{be}\:\mathrm{solved} \\ $$$$\:\:\:\:\:\:\:\mathrm{using}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following}\:``\mathrm{models}''.\:\mathrm{you} \\ $$$$\:\:\:\:\:\:\:\mathrm{have}\:\mathrm{to}\:\mathrm{expand}\:\mathrm{and}\:\mathrm{match}\:\mathrm{the}\:\mathrm{constants} \\ $$$$\:\:\:\:\:\:\:\mathrm{which}\:\mathrm{leads}\:\mathrm{to}\:\mathrm{systems}\:\mathrm{of}\:\mathrm{4}\:\mathrm{equations}\:\mathrm{for} \\ $$$$\:\:\:\:\:\:\:\alpha,\:\beta,\:\gamma,\:\delta.\:\mathrm{again}\:\mathrm{leading}\:\mathrm{to}\:\mathrm{polynomes}\:\mathrm{of} \\ $$$$\:\:\:\:\:\:\:\mathrm{degree}\:\mathrm{3}\:\mathrm{for}\:\left(\alpha^{\mathrm{2}} \right)\:\mathrm{or}\:\left(\gamma^{\mathrm{2}} \right)\:\mathrm{which}\:\mathrm{might}\:\mathrm{have} \\ $$$$\:\:\:\:\:\:\:``\mathrm{nice}''\:\mathrm{solutions} \\ $$$$\:\:\:\:\:\:\:\left({x}−\alpha−\sqrt{\beta}\right)\left({x}−\alpha+\sqrt{\beta}\right)\left({x}−\gamma−\sqrt{\delta}\right)\left({x}−\gamma+\sqrt{\delta}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\left({x}−\alpha−\sqrt{\beta}−\sqrt{\gamma}−\sqrt{\delta}\right)\left({x}−\alpha−\sqrt{\beta}+\sqrt{\gamma}+\sqrt{\delta}\right)\left({x}−\alpha+\sqrt{\beta}−\sqrt{\gamma}+\sqrt{\delta}\right)\left({x}−\alpha+\sqrt{\beta}+\sqrt{\gamma}−\sqrt{\delta}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\left({x}+\alpha+\sqrt{\beta}+\sqrt{\gamma}+\sqrt{\delta}\right)\left({x}+\alpha+\sqrt{\beta}−\sqrt{\gamma}−\sqrt{\delta}\right)\left({x}+\alpha−\sqrt{\beta}+\sqrt{\gamma}−\sqrt{\delta}\right)\left({x}+\alpha−\sqrt{\beta}+>−\sqrt{\gamma}+\sqrt{\delta}\right)=\mathrm{0} \\ $$$$\mathrm{if}\:\mathrm{all}\:\mathrm{these}\:\mathrm{don}'\mathrm{t}\:\mathrm{work}\:\mathrm{you}\:\mathrm{can}\:\mathrm{only}\:\mathrm{approximate} \\ $$$$\mathrm{because}\:\mathrm{Ferrari}'\mathrm{s}\:\mathrm{solution}\:\mathrm{is}\:\mathrm{always}\:\mathrm{possible} \\ $$$$\mathrm{to}\:\mathrm{find}\:\mathrm{but}\:\mathrm{never}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{handle} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com