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the two roots of eq x^2 +2ax−2b^2 =0 are α&β a,b are rational but a^2 +b^2 is not complete square. make a quad eq whose one root is α+β+(√(α^2 +β^2 ))

$${the}\:{two}\:{roots}\:{of}\:{eq}\:{x}^{\mathrm{2}} +\mathrm{2}{ax}−\mathrm{2}{b}^{\mathrm{2}} =\mathrm{0}\:{are} \\ $$$$\alpha\&\beta \\ $$$${a},{b}\:{are}\:{rational}\:{but}\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \:{is}\:{not}\:{complete} \\ $$$${square}.\:{make}\:{a}\:{quad}\:{eq}\:{whose}\:{one}\:{root}\:{is}\: \\ $$$$\alpha+\beta+\sqrt{\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} } \\ $$

Question Number 227607 Answers: 2 Comments: 0

Question Number 227592 Answers: 0 Comments: 8

∀ε>0,∃δ>0 s.t. ∣h∣<δ ⇒ ∣((e^(x_0 +h) −e^x_0 )/h)−e^x_0 ∣<ε... (Don′t use Taylar/Maclaurin Series, and derivation only use inequality properties Damn.....i stuck here again .. e^x_0 ∣((e^h −1)/h)−1∣<𝛆 ..... fuck ∣((e^h −1)/h)−1∣ how can i show that h+1<e^h <(1/(1−h)) , ∣h∣<1

$$\forall\epsilon>\mathrm{0},\exists\delta>\mathrm{0}\:\mathrm{s}.\mathrm{t}.\:\mid{h}\mid<\delta\:\Rightarrow\:\mid\frac{{e}^{{x}_{\mathrm{0}} +{h}} −{e}^{{x}_{\mathrm{0}} } }{{h}}−{e}^{{x}_{\mathrm{0}} } \mid<\epsilon... \\ $$$$\left(\mathrm{Don}'\mathrm{t}\:\mathrm{use}\:\mathrm{Taylar}/\mathrm{Maclaurin}\:\mathrm{Series},\:\mathrm{and}\:\mathrm{derivation}\right. \\ $$$$\mathrm{only}\:\mathrm{use}\:\mathrm{inequality}\:\mathrm{properties} \\ $$$$\mathrm{Damn}.....\mathrm{i}\:\mathrm{stuck}\:\mathrm{here}\:\mathrm{again}\:..\:\: \\ $$$${e}^{{x}_{\mathrm{0}} } \mid\frac{{e}^{{h}} −\mathrm{1}}{{h}}−\mathrm{1}\mid<\boldsymbol{\epsilon}\:\:..... \\ $$$$\mathrm{fuck}\:\mid\frac{{e}^{{h}} −\mathrm{1}}{{h}}−\mathrm{1}\mid\: \\ $$$$\mathrm{how}\:\mathrm{can}\:\mathrm{i}\:\mathrm{show}\:\mathrm{that}\:{h}+\mathrm{1}<{e}^{{h}} <\frac{\mathrm{1}}{\mathrm{1}−{h}}\:,\:\mid{h}\mid<\mathrm{1} \\ $$

Question Number 227582 Answers: 1 Comments: 0

cos^2 (((3x)/2) + ((11π)/2)) - sin^2 (((3x)/2) + ((11π)/2)) = - (1/2) x = ?

$$\mathrm{cos}^{\mathrm{2}} \left(\frac{\mathrm{3x}}{\mathrm{2}}\:+\:\frac{\mathrm{11}\pi}{\mathrm{2}}\right)\:-\:\mathrm{sin}^{\mathrm{2}} \left(\frac{\mathrm{3x}}{\mathrm{2}}\:+\:\frac{\mathrm{11}\pi}{\mathrm{2}}\right)\:=\:-\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{x}\:=\:? \\ $$

Question Number 227580 Answers: 1 Comments: 0

(π/4)arctan(tan((7π)/8))+tan2x=(1/2)cos(arccos(−(1/2))+(π/3)) x = ?

$$\frac{\pi}{\mathrm{4}}\mathrm{arctan}\left(\mathrm{tan}\frac{\mathrm{7}\pi}{\mathrm{8}}\right)+\mathrm{tan2x}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\left(\mathrm{arccos}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)+\frac{\pi}{\mathrm{3}}\right) \\ $$$$\mathrm{x}\:=\:? \\ $$

Question Number 227579 Answers: 1 Comments: 0

cos6x = ((2tan(π/8))/(1−tan^2 (π/8))) ⇒ x = ?

$$\mathrm{cos6x}\:=\:\frac{\mathrm{2tan}\frac{\pi}{\mathrm{8}}}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \frac{\pi}{\mathrm{8}}}\:\:\:\:\:\Rightarrow\:\:\:\:\mathrm{x}\:=\:? \\ $$

Question Number 227578 Answers: 1 Comments: 0

sin3x + sin^2 (π/4) = cos^2 (π/4) + ((√3)/2) x = ?

$$\mathrm{sin3x}\:+\:\mathrm{sin}^{\mathrm{2}} \frac{\pi}{\mathrm{4}}\:=\:\mathrm{cos}^{\mathrm{2}} \frac{\pi}{\mathrm{4}}\:+\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\mathrm{x}\:=\:? \\ $$

Question Number 227567 Answers: 0 Comments: 0

Question Number 227568 Answers: 1 Comments: 1

AB and AC are two plane mirrors ∠BAC=α P is a point on AB and a ray falls on AC at i=γ and after n times of reflection the ray becomes parrallel to AC find γ in terms of α & n

$${AB}\:{and}\:{AC}\:{are}\:{two}\:{plane}\:{mirrors} \\ $$$$\angle{BAC}=\alpha \\ $$$${P}\:{is}\:{a}\:{point}\:{on}\:{AB}\:{and}\:{a}\:{ray}\:{falls}\:{on}\:{AC} \\ $$$${at}\:{i}=\gamma\:{and}\:{after}\:{n}\:{times}\:{of}\:{reflection} \\ $$$${the}\:{ray}\:{becomes}\:{parrallel}\:{to}\:{AC} \\ $$$${find}\:\gamma\:{in}\:{terms}\:\:{of}\:\alpha\:\&\:{n} \\ $$

Question Number 227561 Answers: 1 Comments: 1

Question Number 227534 Answers: 0 Comments: 8

Post some of the hardests problem you think fall within thescope of high school mathp cometitions under this thread. :)

$$\mathrm{Post}\:\mathrm{some}\:\mathrm{of}\:\mathrm{the}\:\mathrm{hardests} \\ $$$$\mathrm{problem}\:\mathrm{you}\:\mathrm{think}\:\mathrm{fall}\:\mathrm{within}\: \\ $$$$\mathrm{thescope}\:\mathrm{of}\:\mathrm{high}\:\mathrm{school}\:\mathrm{mathp} \\ $$$$\mathrm{cometitions}\:\mathrm{under}\:\mathrm{this}\:\mathrm{thread}. \\ $$$$\left.:\right) \\ $$

Question Number 227533 Answers: 2 Comments: 0

Question Number 227531 Answers: 0 Comments: 0

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