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Question Number 111619 by mathdave last updated on 05/Sep/20

solve for x in x^2 +4x=(√(40x^2 +8x−16))

$${solve}\:{for}\:{x}\:{in}\: \\ $$$${x}^{\mathrm{2}} +\mathrm{4}{x}=\sqrt{\mathrm{40}{x}^{\mathrm{2}} +\mathrm{8}{x}−\mathrm{16}} \\ $$

Commented by Her_Majesty last updated on 04/Sep/20

square it, solve the 4^(th) degree polynomial and check the solutions, 1 of the 3 is false exact solutions possible but hard to handle, approximation gives x_1 ≈−10.2894606 x_2 ≈.914622644 x_3 ≈2.16142299

$${square}\:{it},\:{solve}\:{the}\:\mathrm{4}^{{th}} \:{degree}\:{polynomial} \\ $$$${and}\:{check}\:{the}\:{solutions},\:\mathrm{1}\:{of}\:{the}\:\mathrm{3}\:{is}\:{false} \\ $$$${exact}\:{solutions}\:{possible}\:{but}\:{hard}\:{to}\:{handle}, \\ $$$${approximation}\:{gives} \\ $$$${x}_{\mathrm{1}} \approx−\mathrm{10}.\mathrm{2894606} \\ $$$${x}_{\mathrm{2}} \approx.\mathrm{914622644} \\ $$$${x}_{\mathrm{3}} \approx\mathrm{2}.\mathrm{16142299} \\ $$

Answered by mathdave last updated on 04/Sep/20

solution let x^2 +4x=2(√(10x^2 +8x−4)) ........(1) let t^2 =10x^2 +8x−4 then t^2 =2x^2 +8x^2 +8x−4 t^2 −8x^2 +4=2x^2 +8x t^2 −8x^2 +4=2(x^2 +4x).........(2) from (1) x^2 +4x=2t........(x) multiply equation (x) by 2 2(x^2 +4x)=4t..........(3) equating (2) and (3) t^2 −8x^2 +4=4t⇒t^2 −4t−8x^2 +4=0 t^2 −4t+4−4−8x^2 +4=0 (t−2)^2 −8x^2 =0⇒(t−2)^2 −(2(√2)x)^2 (t−2−2(√2)x)(t−2+2(√2)x)=0 t=2+2(√2)x 0r 2−2(√2)x but t^2 =10x^2 +8x−4 (2+2(√2)x)^2 =10x^2 +8x−4 2x^2 +8x(1−(√2))−8=0 x=((−4(1−(√2))±(√(64−32(√2))))/2) ∵x_1 =(2(√2)−2)+2(√(4−2(√2))) and x_2 =(2(√2)−2)−2(√(4−2(√2))) and at t=2−2(√2)x then (2−2(√2)x)=10x^2 +8x−4 x^2 +4(1+(√2))x−4 x_(3,4) =2(−(√2)−1±(√(4+2(√2)))) x_3 =2(−1−(√2)+(√(4+2(√2)))) and x_4 =−2(1+(√2)+(√(4+(√2)))) mathdave

$${solution} \\ $$$${let}\:{x}^{\mathrm{2}} +\mathrm{4}{x}=\mathrm{2}\sqrt{\mathrm{10}{x}^{\mathrm{2}} +\mathrm{8}{x}−\mathrm{4}}\:\:........\left(\mathrm{1}\right) \\ $$$${let}\:{t}^{\mathrm{2}} =\mathrm{10}{x}^{\mathrm{2}} +\mathrm{8}{x}−\mathrm{4} \\ $$$${then}\:\:{t}^{\mathrm{2}} =\mathrm{2}{x}^{\mathrm{2}} +\mathrm{8}{x}^{\mathrm{2}} +\mathrm{8}{x}−\mathrm{4} \\ $$$${t}^{\mathrm{2}} −\mathrm{8}{x}^{\mathrm{2}} +\mathrm{4}=\mathrm{2}{x}^{\mathrm{2}} +\mathrm{8}{x} \\ $$$${t}^{\mathrm{2}} −\mathrm{8}{x}^{\mathrm{2}} +\mathrm{4}=\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{4}{x}\right).........\left(\mathrm{2}\right) \\ $$$${from}\:\left(\mathrm{1}\right) \\ $$$${x}^{\mathrm{2}} +\mathrm{4}{x}=\mathrm{2}{t}........\left({x}\right) \\ $$$${multiply}\:{equation}\:\left({x}\right)\:{by}\:\:\mathrm{2} \\ $$$$\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{4}{x}\right)=\mathrm{4}{t}..........\left(\mathrm{3}\right) \\ $$$${equating}\:\:\left(\mathrm{2}\right)\:{and}\:\left(\mathrm{3}\right) \\ $$$${t}^{\mathrm{2}} −\mathrm{8}{x}^{\mathrm{2}} +\mathrm{4}=\mathrm{4}{t}\Rightarrow{t}^{\mathrm{2}} −\mathrm{4}{t}−\mathrm{8}{x}^{\mathrm{2}} +\mathrm{4}=\mathrm{0} \\ $$$${t}^{\mathrm{2}} −\mathrm{4}{t}+\mathrm{4}−\mathrm{4}−\mathrm{8}{x}^{\mathrm{2}} +\mathrm{4}=\mathrm{0} \\ $$$$\left({t}−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{8}{x}^{\mathrm{2}} =\mathrm{0}\Rightarrow\left({t}−\mathrm{2}\right)^{\mathrm{2}} −\left(\mathrm{2}\sqrt{\mathrm{2}}{x}\right)^{\mathrm{2}} \\ $$$$\left({t}−\mathrm{2}−\mathrm{2}\sqrt{\mathrm{2}}{x}\right)\left({t}−\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}{x}\right)=\mathrm{0} \\ $$$${t}=\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}{x}\:\mathrm{0}{r}\:\mathrm{2}−\mathrm{2}\sqrt{\mathrm{2}}{x} \\ $$$${but}\:{t}^{\mathrm{2}} =\mathrm{10}{x}^{\mathrm{2}} +\mathrm{8}{x}−\mathrm{4} \\ $$$$\left(\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}{x}\right)^{\mathrm{2}} =\mathrm{10}{x}^{\mathrm{2}} +\mathrm{8}{x}−\mathrm{4} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} +\mathrm{8}{x}\left(\mathrm{1}−\sqrt{\mathrm{2}}\right)−\mathrm{8}=\mathrm{0} \\ $$$${x}=\frac{−\mathrm{4}\left(\mathrm{1}−\sqrt{\mathrm{2}}\right)\pm\sqrt{\mathrm{64}−\mathrm{32}\sqrt{\mathrm{2}}}}{\mathrm{2}} \\ $$$$\because{x}_{\mathrm{1}} =\left(\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{2}\right)+\mathrm{2}\sqrt{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}}\:\:{and}\:{x}_{\mathrm{2}} =\left(\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{2}\right)−\mathrm{2}\sqrt{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$${and}\:{at}\:{t}=\mathrm{2}−\mathrm{2}\sqrt{\mathrm{2}}{x} \\ $$$${then}\:\:\:\left(\mathrm{2}−\mathrm{2}\sqrt{\mathrm{2}}{x}\right)=\mathrm{10}{x}^{\mathrm{2}} +\mathrm{8}{x}−\mathrm{4} \\ $$$${x}^{\mathrm{2}} +\mathrm{4}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right){x}−\mathrm{4} \\ $$$${x}_{\mathrm{3},\mathrm{4}} =\mathrm{2}\left(−\sqrt{\mathrm{2}}−\mathrm{1}\pm\sqrt{\mathrm{4}+\mathrm{2}\sqrt{\mathrm{2}}}\right) \\ $$$${x}_{\mathrm{3}} =\mathrm{2}\left(−\mathrm{1}−\sqrt{\mathrm{2}}+\sqrt{\mathrm{4}+\mathrm{2}\sqrt{\mathrm{2}}}\right)\:\:\:{and}\:\:{x}_{\mathrm{4}} =−\mathrm{2}\left(\mathrm{1}+\sqrt{\mathrm{2}}+\sqrt{\mathrm{4}+\sqrt{\mathrm{2}}}\right) \\ $$$${mathdave} \\ $$$$ \\ $$

Commented by Her_Majesty last updated on 04/Sep/20

but the question you posted is x^2 +4x=(√(40x^2 +3x−16)) and you can′t solve it this way.

$${but}\:{the}\:{question}\:{you}\:{posted}\:{is} \\ $$$${x}^{\mathrm{2}} +\mathrm{4}{x}=\sqrt{\mathrm{40}{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{16}} \\ $$$${and}\:{you}\:{can}'{t}\:{solve}\:{it}\:{this}\:{way}. \\ $$

Commented by Her_Majesty last updated on 04/Sep/20

anyway one of your 4 solutions doesn′t solve the equation x^2 +4x=(√(40x^2 +32x−16)), so please finish the work

$${anyway}\:{one}\:{of}\:{your}\:\mathrm{4}\:{solutions}\:{doesn}'{t}\:{solve} \\ $$$${the}\:{equation}\:{x}^{\mathrm{2}} +\mathrm{4}{x}=\sqrt{\mathrm{40}{x}^{\mathrm{2}} +\mathrm{32}{x}−\mathrm{16}},\:{so} \\ $$$${please}\:{finish}\:{the}\:{work} \\ $$

Commented by Her_Majesty last updated on 04/Sep/20

x^2 +4x=(√(40x^2 +32x−16)) squaring and transforming x^4 +8x^3 −24x−32x+16=0 (x^2 +4(1+(√2))x−4)(x^2 +4(1−(√2))x−4)=0 x_1 =−2(1+(√2)+(√(4+2(√2)))) x_2 =−2(1+(√2)−(√(4+2(√2)))) x_3 =−2(1−(√2)+(√(4−2(√2)))) [false!] x_4 =−2(1−(√2)−(√(4−2(√2))))

$${x}^{\mathrm{2}} +\mathrm{4}{x}=\sqrt{\mathrm{40}{x}^{\mathrm{2}} +\mathrm{32}{x}−\mathrm{16}} \\ $$$${squaring}\:{and}\:{transforming} \\ $$$${x}^{\mathrm{4}} +\mathrm{8}{x}^{\mathrm{3}} −\mathrm{24}{x}−\mathrm{32}{x}+\mathrm{16}=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} +\mathrm{4}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right){x}−\mathrm{4}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\left(\mathrm{1}−\sqrt{\mathrm{2}}\right){x}−\mathrm{4}\right)=\mathrm{0} \\ $$$${x}_{\mathrm{1}} =−\mathrm{2}\left(\mathrm{1}+\sqrt{\mathrm{2}}+\sqrt{\mathrm{4}+\mathrm{2}\sqrt{\mathrm{2}}}\right) \\ $$$${x}_{\mathrm{2}} =−\mathrm{2}\left(\mathrm{1}+\sqrt{\mathrm{2}}−\sqrt{\mathrm{4}+\mathrm{2}\sqrt{\mathrm{2}}}\right) \\ $$$${x}_{\mathrm{3}} =−\mathrm{2}\left(\mathrm{1}−\sqrt{\mathrm{2}}+\sqrt{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}}\right)\:\left[{false}!\right] \\ $$$${x}_{\mathrm{4}} =−\mathrm{2}\left(\mathrm{1}−\sqrt{\mathrm{2}}−\sqrt{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}}\right) \\ $$

Commented by Her_Majesty last updated on 05/Sep/20

you changed the question now but you changed it to a different from the one you solved... you have to be more careful

$${you}\:{changed}\:{the}\:{question}\:{now}\:{but}\:{you} \\ $$$${changed}\:{it}\:{to}\:{a}\:{different}\:{from}\:{the}\:{one}\:{you} \\ $$$${solved}...\:{you}\:{have}\:{to}\:{be}\:{more}\:{careful} \\ $$

Commented by mathdave last updated on 05/Sep/20

since u discovered that one of the solutiondidnt satisfy d problem then go ahead an fix it

$${since}\:{u}\:{discovered}\:{that}\:{one}\:{of}\:{the} \\ $$$${solutiondidnt}\:{satisfy}\:{d}\:{problem}\:{then} \\ $$$${go}\:{ahead}\:{an}\:{fix}\:{it}\: \\ $$

Commented by mathdave last updated on 05/Sep/20

stop murmuring

$${stop}\:{murmuring}\: \\ $$

Commented by mathdave last updated on 05/Sep/20

more careful to which way, are u my teacher that u are cautioning me

$${more}\:\:{careful}\:{to}\:\:{which}\:{way},\:{are}\:{u}\:{my}\:{teacher}\:{that}\:{u}\:{are}\:{cautioning}\:{me} \\ $$$$ \\ $$

Commented by mathdave last updated on 05/Sep/20

the only way i can believe dat u know mathematics is when u discovered err and do some working on ur own to avert the err but u saying it orally or using a calculator or checking wolfram to determine the answer do nt tell u know mathematics

$${the}\:{only}\:{way}\:{i}\:{can}\:{believe}\:{dat}\:{u}\:{know} \\ $$$${mathematics}\:{is}\:{when}\:{u}\:{discovered}\:{err} \\ $$$${and}\:{do}\:{some}\:{working}\:{on}\:{ur}\:{own}\:{to}\:{avert}\:{the}\:{err} \\ $$$${but}\:{u}\:{saying}\:{it}\:{orally}\:{or}\:{using}\:{a}\:{calculator}\:{or}\:{checking} \\ $$$${wolfram}\:{to}\:{determine}\:{the}\:{answer}\:{do}\:{nt}\:{tell}\:{u}\:{know} \\ $$$${mathematics} \\ $$$$ \\ $$

Commented by Her_Majesty last updated on 05/Sep/20

the point is here, you posted a question that I solved, then you solved a different question, then you changed the question to a third version.

$${the}\:{point}\:{is}\:{here},\:{you}\:{posted}\:{a}\:{question} \\ $$$${that}\:{I}\:{solved},\:{then}\:{you}\:{solved}\:{a}\:{different} \\ $$$${question},\:{then}\:{you}\:{changed}\:{the}\:{question} \\ $$$${to}\:{a}\:{third}\:{version}. \\ $$

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