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Question Number 69995 by behi83417@gmail.com last updated on 29/Sep/19

1. { (((√(x^2 +y^2 ))+(√(x^2 −y^2 ))=a)),(((√(x+y))+(√(x−y))=b)) :} [a,b∈R^+ ] 2. { (((√((√x)+y))+(√(x+(√y)))=a)),(((√((√x)−y))+(√(x−(√y)))=b)) :} 3. { ((x^2 +y^2 =(a−b)xy)),((x^3 +y^3 =ab(x−y))) :} 4. { ((x+y+z=a)),((x^2 +y^2 +z^2 =b)),((x^3 +y^3 +z^3 =abxyz)) :} 5. (√(x^2 +(√x)))+(√(x^2 −(√x)))=2 6. (√(x^2 +x+1))+(√(x^2 +x−1))+(√(x^2 −x−1))=1 7. 16^(sin^2 x) +16^(cos^2 x) =10 8. 2^(lnx) =x.ln(x+(√2))

$$\:\:\mathrm{1}.\begin{cases}{\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{y}}^{\mathrm{2}} }+\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{y}}^{\mathrm{2}} }=\boldsymbol{\mathrm{a}}}\\{\sqrt{\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}}+\sqrt{\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{y}}}=\boldsymbol{\mathrm{b}}}\end{cases}\:\:\:\:\:\left[\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{b}}\in\boldsymbol{\mathrm{R}}^{+} \right] \\ $$$$ \\ $$$$\:\:\:\mathrm{2}.\begin{cases}{\sqrt{\sqrt{\mathrm{x}}+\mathrm{y}}+\sqrt{\mathrm{x}+\sqrt{\mathrm{y}}}=\boldsymbol{\mathrm{a}}}\\{\sqrt{\sqrt{\mathrm{x}}−\mathrm{y}}+\sqrt{\mathrm{x}−\sqrt{\mathrm{y}}}=\boldsymbol{\mathrm{b}}}\end{cases} \\ $$$$\: \\ $$$$\:\:\:\mathrm{3}.\begin{cases}{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{y}}^{\mathrm{2}} =\left(\boldsymbol{\mathrm{a}}−\boldsymbol{\mathrm{b}}\right)\boldsymbol{\mathrm{xy}}}\\{\boldsymbol{\mathrm{x}}^{\mathrm{3}} +\boldsymbol{\mathrm{y}}^{\mathrm{3}} =\boldsymbol{\mathrm{ab}}\left(\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{y}}\right)}\end{cases} \\ $$$$ \\ $$$$\:\:\mathrm{4}.\begin{cases}{\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}+\boldsymbol{\mathrm{z}}=\boldsymbol{\mathrm{a}}}\\{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{y}}^{\mathrm{2}} +\boldsymbol{\mathrm{z}}^{\mathrm{2}} =\boldsymbol{\mathrm{b}}}\\{\boldsymbol{\mathrm{x}}^{\mathrm{3}} +\boldsymbol{\mathrm{y}}^{\mathrm{3}} +\boldsymbol{\mathrm{z}}^{\mathrm{3}} =\boldsymbol{\mathrm{abxyz}}}\end{cases} \\ $$$$\:\:\:\mathrm{5}.\:\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\sqrt{\boldsymbol{\mathrm{x}}}}+\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\sqrt{\boldsymbol{\mathrm{x}}}}=\mathrm{2} \\ $$$$ \\ $$$$\:\:\:\:\mathrm{6}.\:\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{x}}+\mathrm{1}}+\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{x}}−\mathrm{1}}+\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{x}}−\mathrm{1}}=\mathrm{1} \\ $$$$ \\ $$$$\:\mathrm{7}.\:\:\mathrm{16}^{\boldsymbol{\mathrm{sin}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}} +\mathrm{16}^{\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}} =\mathrm{10} \\ $$$$ \\ $$$$\:\:\mathrm{8}.\:\:\:\:\:\:\:\:\:\:\mathrm{2}^{\boldsymbol{\mathrm{lnx}}} =\boldsymbol{\mathrm{x}}.\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}+\sqrt{\mathrm{2}}\right) \\ $$

Answered by mr W last updated on 29/Sep/19

Q7: let t=16^(sin^2 x) 16^(cos^2 x) =16^(1−sin^2 x) =((16)/(16^(sin^2 x) ))=((16)/t) t+((16)/t)=10 ⇒t^2 −10t+16=0 (t−2)(t−8)=0 ⇒t=2 or 8 16^(sin^2 x) =2 ⇒sin^2 x=((ln 2)/(ln 16))=(1/4) ⇒sin x=±(1/2) ⇒x=nπ±(π/6) 16^(sin^2 x) =8 ⇒sin^2 x=((ln 8)/(ln 16))=(3/4) ⇒sin x=±((√3)/2) ⇒x=nπ±(π/3)

$${Q}\mathrm{7}: \\ $$$${let}\:{t}=\mathrm{16}^{\mathrm{sin}^{\mathrm{2}} \:{x}} \\ $$$$\mathrm{16}^{\mathrm{cos}^{\mathrm{2}} \:{x}} =\mathrm{16}^{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:{x}} =\frac{\mathrm{16}}{\mathrm{16}^{\mathrm{sin}^{\mathrm{2}} \:{x}} }=\frac{\mathrm{16}}{{t}} \\ $$$${t}+\frac{\mathrm{16}}{{t}}=\mathrm{10} \\ $$$$\Rightarrow{t}^{\mathrm{2}} −\mathrm{10}{t}+\mathrm{16}=\mathrm{0} \\ $$$$\left({t}−\mathrm{2}\right)\left({t}−\mathrm{8}\right)=\mathrm{0} \\ $$$$\Rightarrow{t}=\mathrm{2}\:{or}\:\mathrm{8} \\ $$$$ \\ $$$$\mathrm{16}^{\mathrm{sin}^{\mathrm{2}} \:{x}} =\mathrm{2} \\ $$$$\Rightarrow\mathrm{sin}^{\mathrm{2}} \:{x}=\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{ln}\:\mathrm{16}}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{sin}\:{x}=\pm\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{x}={n}\pi\pm\frac{\pi}{\mathrm{6}} \\ $$$$\mathrm{16}^{\mathrm{sin}^{\mathrm{2}} \:{x}} =\mathrm{8} \\ $$$$\Rightarrow\mathrm{sin}^{\mathrm{2}} \:{x}=\frac{\mathrm{ln}\:\mathrm{8}}{\mathrm{ln}\:\mathrm{16}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{sin}\:{x}=\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\Rightarrow{x}={n}\pi\pm\frac{\pi}{\mathrm{3}} \\ $$

Commented by behi83417@gmail.com last updated on 29/Sep/19

thanks in advance dear master. #7 , is so easy for you. please try anothers! (ha ha excuse me)

$$\mathrm{thanks}\:\mathrm{in}\:\mathrm{advance}\:\mathrm{dear}\:\mathrm{master}. \\ $$$$#\mathrm{7}\:,\:\mathrm{is}\:\mathrm{so}\:\mathrm{easy}\:\mathrm{for}\:\mathrm{you}. \\ $$$$\mathrm{please}\:\mathrm{try}\:\mathrm{anothers}!\:\:\left(\mathrm{ha}\:\mathrm{ha}\:\mathrm{excuse}\:\mathrm{me}\right) \\ $$

Answered by mr W last updated on 29/Sep/19

Q5: 2x^2 +2(√(x^4 −x))=4 (√(x^4 −x))=2−x^2 x^4 −x=x^4 −4x^2 +4 4x^2 −x−4=0 ⇒x=((1+(√(65)))/8)

$${Q}\mathrm{5}: \\ $$$$\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}\sqrt{{x}^{\mathrm{4}} −{x}}=\mathrm{4} \\ $$$$\sqrt{{x}^{\mathrm{4}} −{x}}=\mathrm{2}−{x}^{\mathrm{2}} \\ $$$${x}^{\mathrm{4}} −{x}={x}^{\mathrm{4}} −\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} −{x}−\mathrm{4}=\mathrm{0} \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}+\sqrt{\mathrm{65}}}{\mathrm{8}} \\ $$

Answered by ajfour last updated on 29/Sep/19

(1) { (((√(x^2 +y^2 ))+(√(x^2 −y^2 ))=a)),(((√(x+y))+(√(x−y))=b)) :} let y=tx , ∣x∣=r (√(1+t^2 ))+(√(1−t^2 ))=(a/r) (√(1+t))+(√(1−t))=(b/(√r)) (√(1+t^2 ))+(√(1−t^2 ))=((2a)/b^2 )(1+(√(1−t^2 )) ) let t^2 =cos 2θ , ((2a)/b^2 )=μ (√2)cos θ+(√2)sin θ=(((2a)/b^2 ))(1+(√2)sin θ) cos θ+(1−μ)sin θ=(μ/(√2)) cos {φ−θ}=(μ/((√2)(√(1+(1−μ)^2 )))) φ=tan^(−1) (1−μ) =tan^(−1) (1−((2a)/b^2 )) t^2 =cos 2θ θ = φ−cos^(−1) (μ/((√2)(√(1+(1−μ)^2 )))) x=r = (a/((√(1+t^2 ))+(√(1−t^2 )))) y=tx .

$$\left(\mathrm{1}\right)\:\:\:\begin{cases}{\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{y}}^{\mathrm{2}} }+\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{y}}^{\mathrm{2}} }=\boldsymbol{\mathrm{a}}}\\{\sqrt{\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}}+\sqrt{\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{y}}}=\boldsymbol{\mathrm{b}}}\end{cases} \\ $$$${let}\:\:{y}={tx}\:,\:\mid{x}\mid={r} \\ $$$$\:\:\:\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }+\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }=\frac{{a}}{{r}} \\ $$$$\:\:\:\sqrt{\mathrm{1}+{t}}+\sqrt{\mathrm{1}−{t}}=\frac{{b}}{\sqrt{{r}}} \\ $$$$\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }+\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }=\frac{\mathrm{2}{a}}{{b}^{\mathrm{2}} }\left(\mathrm{1}+\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:\right) \\ $$$${let}\:{t}^{\mathrm{2}} =\mathrm{cos}\:\mathrm{2}\theta\:\:,\:\frac{\mathrm{2}{a}}{{b}^{\mathrm{2}} }=\mu \\ $$$$\:\sqrt{\mathrm{2}}\mathrm{cos}\:\theta+\sqrt{\mathrm{2}}\mathrm{sin}\:\theta=\left(\frac{\mathrm{2}{a}}{{b}^{\mathrm{2}} }\right)\left(\mathrm{1}+\sqrt{\mathrm{2}}\mathrm{sin}\:\theta\right) \\ $$$$\mathrm{cos}\:\theta+\left(\mathrm{1}−\mu\right)\mathrm{sin}\:\theta=\frac{\mu}{\sqrt{\mathrm{2}}} \\ $$$$\mathrm{cos}\:\left\{\phi−\theta\right\}=\frac{\mu}{\sqrt{\mathrm{2}}\sqrt{\mathrm{1}+\left(\mathrm{1}−\mu\right)^{\mathrm{2}} }} \\ $$$$\:\:\phi=\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{1}−\mu\right)\:=\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{1}−\frac{\mathrm{2}{a}}{{b}^{\mathrm{2}} }\right) \\ $$$${t}^{\mathrm{2}} =\mathrm{cos}\:\mathrm{2}\theta \\ $$$$\:\:\:\:\theta\:=\:\phi−\mathrm{cos}^{−\mathrm{1}} \frac{\mu}{\sqrt{\mathrm{2}}\sqrt{\mathrm{1}+\left(\mathrm{1}−\mu\right)^{\mathrm{2}} }} \\ $$$$\:\:{x}={r}\:=\:\frac{{a}}{\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }+\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }} \\ $$$$\:\:{y}={tx}\:. \\ $$

Commented by behi83417@gmail.com last updated on 30/Sep/19

thank you very much sir Ajfour. you have creative mind,questions and solutions.god bless you sir.

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{sir}\:\mathrm{Ajfour}. \\ $$$$\mathrm{you}\:\mathrm{have}\:\mathrm{creative}\:\mathrm{mind},\mathrm{questions}\:\mathrm{and} \\ $$$$\mathrm{solutions}.\mathrm{god}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Answered by mr W last updated on 01/Oct/19

(3) x+y=u xy=v (x+y)^2 −2xy=(a−b)xy u^2 =(a−b+2)v x^3 +y^3 =(x+y)[(x+y)^2 −3xy]=u(u^2 −3v) x−y=(√((x+y)^2 −4xy))=(√(u^2 −4v)) u(u^2 −3v)=ab(√(u^2 −4v)) uv(a−b−1)=ab(√(u^2 −4v)) u^2 v^2 (a−b−1)^2 =a^2 b^2 (u^2 −4v) (a−b+2)(a−b−1)^2 v^2 =a^2 b^2 (a−b−2) ⇒v=((ab)/(a−b−1))(√((a−b−2)/(a−b+2)))=p say ⇒u=±(√((ab(√((a−b+2)(a−b−2))))/(a−b−1)))=±q say x^2 ±qx+p=0 x,y=(1/2)(±q±(√(q^2 −4p)))

$$\left(\mathrm{3}\right) \\ $$$${x}+{y}={u} \\ $$$${xy}={v} \\ $$$$\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{2}{xy}=\left({a}−{b}\right){xy} \\ $$$${u}^{\mathrm{2}} =\left({a}−{b}+\mathrm{2}\right){v} \\ $$$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} =\left({x}+{y}\right)\left[\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{3}{xy}\right]={u}\left({u}^{\mathrm{2}} −\mathrm{3}{v}\right) \\ $$$${x}−{y}=\sqrt{\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{4}{xy}}=\sqrt{{u}^{\mathrm{2}} −\mathrm{4}{v}} \\ $$$${u}\left({u}^{\mathrm{2}} −\mathrm{3}{v}\right)={ab}\sqrt{{u}^{\mathrm{2}} −\mathrm{4}{v}} \\ $$$${uv}\left({a}−{b}−\mathrm{1}\right)={ab}\sqrt{{u}^{\mathrm{2}} −\mathrm{4}{v}} \\ $$$${u}^{\mathrm{2}} {v}^{\mathrm{2}} \left({a}−{b}−\mathrm{1}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} {b}^{\mathrm{2}} \left({u}^{\mathrm{2}} −\mathrm{4}{v}\right) \\ $$$$\left({a}−{b}+\mathrm{2}\right)\left({a}−{b}−\mathrm{1}\right)^{\mathrm{2}} {v}^{\mathrm{2}} ={a}^{\mathrm{2}} {b}^{\mathrm{2}} \left({a}−{b}−\mathrm{2}\right) \\ $$$$\Rightarrow{v}=\frac{{ab}}{{a}−{b}−\mathrm{1}}\sqrt{\frac{{a}−{b}−\mathrm{2}}{{a}−{b}+\mathrm{2}}}={p}\:{say} \\ $$$$\Rightarrow{u}=\pm\sqrt{\frac{{ab}\sqrt{\left({a}−{b}+\mathrm{2}\right)\left({a}−{b}−\mathrm{2}\right)}}{{a}−{b}−\mathrm{1}}}=\pm{q}\:{say} \\ $$$${x}^{\mathrm{2}} \pm{qx}+{p}=\mathrm{0} \\ $$$${x},{y}=\frac{\mathrm{1}}{\mathrm{2}}\left(\pm{q}\pm\sqrt{{q}^{\mathrm{2}} −\mathrm{4}{p}}\right) \\ $$

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