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Question Number 50511 by behi83417@gmail.com last updated on 17/Dec/18

$$\text{You can't use 'macro parameter character \#' in math mode}$$$$\sqrt{a+\sqrt{a-x}}+\sqrt{a-\sqrt{a+x}}=2x$$

Answered by behi83417@gmail.com last updated on 17/Dec/18

Commented by behi83417@gmail.com last updated on 17/Dec/18

$$if\mathit{a}=0,then:\mathit{x}=0canacceptasanswer.$$

Commented by behi83417@gmail.com last updated on 17/Dec/18

Commented by behi83417@gmail.com last updated on 17/Dec/18

Commented by behi83417@gmail.com last updated on 17/Dec/18

Commented by mr W last updated on 17/Dec/18

$$canyougivethesolutionfora$$$$concretevalueofa,e.g.a=20?$$

Commented by behi83417@gmail.com last updated on 17/Dec/18

$$excusemesir.icantundrestandthis$$$$commenet.[concrete=?]$$

Commented by mr W last updated on 17/Dec/18

$$Imeanyoutakeaspecialvalueforthe$$$$parametera,saya=20,andshow$$$$yoursolutionforthisspecialcase.$$$$i.e.$$$$\sqrt{20+\sqrt{20-x}}+\sqrt{20-\sqrt{20+x}}=2x$$

Commented by behi83417@gmail.com last updated on 18/Dec/18

$$a=20,istoolargeandbignumbersin$$$$solution.solvefor:a=1$$$$\sqrt{1+\sqrt{1-x}}+\sqrt{1-\sqrt{1+x}}=2x$$$$1+x=t^2,1-x=s^2\Rightarrow x=\frac{t^2-s^2}{2},t^2+s^2=2$$$$\sqrt{1+s}+\sqrt{1-t}=t^2-s^2$$$$1+s+1-t+2\sqrt{(1+s)(1-t)}={(t^2-s^2)}^2$$$$4(1+s)(1-t)={[{(t^2-s^2)}^2+t-s-2]}^2$$$$4-4t+4s-4ts={[t^4+s^4-2t^2{s}^2+t-s-2]}^2=$$$$=t^8+s^8+4t^4{s}^4+t^2+s^2+4+2t^4{s}^4-4t^6{s}^2$$$$+2t^5-2t^{4}s-4t^4-4t^2{s}^6+2{ts}^4-2s^5-4s^4$$$$-4t^3{s}^2+4t^2{s}^3+4t^2{s}^2-4t+4s-2ts$$$$(t^8+s^8)+2(t^5-s^5)+6t^4{s}^4+2ts+4t^2{s}^2$$$$-4t^2{s}^2(t^4+s^4)-2ts(t^3-s^3)+4(t^4+s^4)$$$$-4t^2{s}^2(t-s)=0$$$$(t^8+s^8)+2(t^5-s^5)-4(1+t^2{s}^2)(t^4+s^4)$$$$-2ts(t^3-s^3)-4t^2{s}^2(t-s)+6t^4{s}^4+$$$$+8t^2{s}^2+2ts+(t^2+s^2)=0$$$$let:\mathbf{t}-\mathbf{s}=\mathbf{p},\mathbf{ts}=\mathbf{q}$$$$t^8+s^8=p^8+8p^{6}q+20p^4{q}^2+16p^2{q}^3+2q^4$$$$2(t^5-s^5)=2p^5+10p^{2}q+10{pq}^2$$$$4(1-t^2{s}^2)(t^4+s^4)=4(p^4+4p^{2}q+2q^2)(1-q^2)=$$$$=4p^4+16p^{2}q+8q^2-4p^4{q}^2-16p^2{q}^3-8q^4$$$$-2ts(t^3-s^3)=-2q(p^3+3pq)=$$$$=-2p^{3}q-6{pq}^2$$$$-4t^2{s}^2(t-s)=-4q^{2}p$$$$-2p^{3}q-10{pq}^2+2q+4p^4=0$$$$16p^4{q}^2+2(4p^6+4p^3-8p^2+2)q$$$$+(p^8+2p^5-4p^4+p^2)=0$$$${△}^{\prime }={(4p^6+4p^3-8p^2+2)}^2-16p^4(p^8+2p^5-4p^4+p^2)=$$$$=4{(2p^3-4p^2+1)}^2$$$$q=\frac{-(4p^6+4p^3-8p^2+2)\pm 2(2p^3-4p^2+1)}{16p^4}=$$$$q_1=\frac{-4p^6}{16p^4}=-\frac{p^2}{4}$$$$q_2=-\frac{p^6+2p^3-4p^2+1}{4p^4}$$$$for:q=-\frac{p^2}{4}\Rightarrow p^2+4q=0\Rightarrow {(t-s)}^2+4ts=0$$$$\Rightarrow {(t+s)}^2=0\Rightarrow t=-s\Rightarrow x=0(notok!)$$$$q\left(p\right)=-\frac{p^6+2p^3-4p^2+1}{4p^4},haveextrimums$$$$atpoints:E(-.64,1.63),F(.75,.18)$$$$p=-.64\Rightarrow t-s=-0.64\Rightarrow t=s-0.64$$$$q=1.63\Rightarrow ts=1.63\Rightarrow s(s-.64)=1.63$$$$\Rightarrow s^2-0.64s-1.63=0\Rightarrow s=\frac{.64\pm 2.63}{2}$$$$\Rightarrow s=1.64,-1\Rightarrow t=1,-1.64$$$$\Rightarrow x=\frac{t^2-s^2}{2}=\frac{1.{64}^2-1}{2}=\pm 0.84$$$$\sqrt{1+\sqrt{1-0.84}}+\sqrt{1-\sqrt{1+0.84}}=$$$$=1.18+0.06i\ne 2\times 0.84$$$$\sqrt{1+\sqrt{1+.84}}+\sqrt{1-\sqrt{1-.84}}\ne 2(-.84)$$$$\Rightarrow x=\pm 0.84notanswer.$$$$p=0.75\Rightarrow t-s=0.75\Rightarrow t=s+0.75$$$$ts=.18\Rightarrow s(s+.75)=.18$$$$\Rightarrow s^2+.75s-.18=0\Rightarrow s=\frac{-.75\pm 1.13}{2}$$$$\Rightarrow s=-.94,0.19,t=-.19,0.94$$$$x=\frac{{(.94)}^2-{(.19)}^2}{2}=\pm 0.42$$$$\sqrt{1+\sqrt{1-.42}}+\sqrt{1-\sqrt{1+.42}}\ne 2\times (.42)$$$$\sqrt{1+\sqrt{1+.42}}+\sqrt{1-\sqrt{1-.42}}\ne 2(-.42)$$$$\Rightarrow x=\pm .42notanswer.$$

Commented by mr W last updated on 18/Dec/18

$$whatisthesolution?ithinkfora=1$$$$thereisnosolution.$$

Commented by behi83417@gmail.com last updated on 18/Dec/18

$$for:a=1,noanswerexcist.$$

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