Respuesta :
Answer:
x=1
Step-by-step explanation:
[tex]\sqrt{3x-1}+\sqrt{5x-3}+\sqrt{x-1}=2\sqrt{2}[/tex]
[tex]\sqrt{3x-1}+\sqrt{5x-3}+\sqrt{x-1}-\sqrt{x-1}=2\sqrt{2}-\sqrt{x-1}[/tex]
[tex]\sqrt{3x-1}+\sqrt{5x-3}=2\sqrt{2}-\sqrt{x-1}[/tex]
Squaring on both sides we get
[tex]\left(\sqrt{3x-1}+\sqrt{5x-3}\right)^2=\left(2\sqrt{2}-\sqrt{x-1}\right)^2[/tex] -----(A)
Expanding Left hand side we get
[tex]\left(\sqrt{3x-1}+\sqrt{5x-3}\right)^2=\left(2\sqrt{2}-\sqrt{x-1}\right)^2[/tex]
=[tex]\left(\sqrt{3x-1}\right)^2+2\sqrt{3x-1}\sqrt{5x-3}+\left(\sqrt{5x-3}\right)^2[/tex]
=[tex]\left(3x-1\right)+2\sqrt{3x-1}\sqrt{5x-3}+\left(5x-3\right)[/tex]
=[tex]8x+2\sqrt{3x-1}\sqrt{5x-3}-4[/tex]
Expanding Right Hand side
[tex]\left(2\sqrt{2}-\sqrt{x-1}\right)^2[/tex]
= [tex]\left(2\sqrt{2}\right)^2-2\cdot \:2\sqrt{2}\sqrt{x-1}+\left(\sqrt{x-1}\right)^2[/tex]
= [tex]8-4\sqrt{2}\sqrt{x-1}+\left(x-1\right)[/tex]
= [tex]x+7-4\sqrt{2}\sqrt{x-1}[/tex]
Hence we have now (A) equal to
[tex]8x+2\sqrt{3x-1}\sqrt{5x-3}-4=x+7-4\sqrt{2}\sqrt{x-1}[/tex]
subtracting 8x from both sides
[tex]8x+2\sqrt{3x-1}\sqrt{5x-3}-4-8x=x+7-4\sqrt{2}\sqrt{x-1}-8x[/tex]
[tex]2\sqrt{3x-1}\sqrt{5x-3}-4=-7x-4\sqrt{2}\sqrt{x-1}+7[/tex]
Adding 4 on both sides we get...
[tex]2\sqrt{3x-1}\sqrt{5x-3}-4+4=-7x-4\sqrt{2}\sqrt{x-1}+7+4[/tex]
[tex]2\sqrt{3x-1}\sqrt{5x-3}=-7x+11-4\sqrt{2}\sqrt{x-1}[/tex]
now squaring on both sides again
[tex]\left(2\sqrt{3x-1}\sqrt{5x-3}\right)^2=\left(-7x+11-4\sqrt{2}\sqrt{x-1}\right)^2[/tex]
----------(B)
Left hand side:
[tex]\left(2\sqrt{3x-1}\sqrt{5x-3}\right)^2[/tex]
= [tex]4\left(3x-1\right)\left(5x-3\right)[/tex]
= [tex]60x^2-56x+12[/tex]
Right hand side:
[tex]\left(-7x+11-4\sqrt{2}\sqrt{x-1}\right)^2[/tex]
= [tex]\left(-7x+11-4\sqrt{2}\sqrt{x-1}\right)\left(-7x+11-4\sqrt{2}\sqrt{x-1}\right)[/tex]
= [tex]49x^2-154x+56\sqrt{2}\sqrt{x-1}x+121-88\sqrt{2}\sqrt{x-1}+32\left(x-1\right)[/tex]
= [tex]49x^2+56\sqrt{2}x\sqrt{x-1}-122x-88\sqrt{2}\sqrt{x-1}+89[/tex]
Hence we have (B) equals to
[tex]60x^2-56x+12=49x^2+56\sqrt{2}x\sqrt{x-1}-122x-88\sqrt{2}\sqrt{x-1}+89[/tex]
[tex]49x^2+56\sqrt{2}x\sqrt{x-1}-122x-88\sqrt{2}\sqrt{x-1}+89=60x^2-56x+12[/tex]
Subtract [tex]49x^2-122x[/tex] from both sides
[tex]56\sqrt{2}x\sqrt{x-1}-88\sqrt{2}\sqrt{x-1}+89=11x^2+66x+12[/tex]
[tex]56\sqrt{2}x\sqrt{x-1}-88\sqrt{2}\sqrt{x-1}=11x^2+66x-77[/tex] ---(C)
Also
[tex]56\sqrt{2}x\sqrt{x-1}-88\sqrt{2}\sqrt{x-1}[/tex]
= [tex]7\cdot \:8\sqrt{x-1}\sqrt{2}x-11\cdot \:8\sqrt{x-1}\sqrt{2}[/tex]
= [tex]8\sqrt{x-1}\sqrt{2}\left(7x-11\right)[/tex]
Hence (C) becomes
[tex]8\sqrt{2}\sqrt{x-1}\left(7x-11\right)=11x^2+66x-77[/tex]
Squaring both sides
[tex]\left(8\sqrt{2}\sqrt{x-1}\left(7x-11\right)\right)^2=\left(11x^2+66x-77\right)^2[/tex]
-------(D)
Left Hand Side
= [tex]6272\left(x-1\right)x^2-19712\left(x-1\right)x+15488\left(x-1\right)[/tex]
= [tex]6272x^3-25984x^2+35200x-15488[/tex]
Right hand Side
= [tex]\left(11x^2+66x-77\right)\left(11x^2+66x-77\right)[/tex]
= [tex]121x^4+1452x^3+2662x^2-10164x+5929[/tex]
Hence (D) becomes
[tex]6272x^3-25984x^2+35200x-15488=121x^4+1452x^3+2662x^2-10164x+5929[/tex]
[tex]121x^4+1452x^3+2662x^2-10164x+5929=6272x^3-25984x^2+35200x-15488[/tex]
subtracting [tex]6272x^3-25984x^2+35200x-15488[/tex] from both sides
[tex]121x^4-4820x^3+28646x^2-45364x+21417=0[/tex] -----(E)
Solving using factoring and trying for all the possible rational roots starting from x=1 , we get it satisfies at x=1 itself. Hence (x-1) is one of the factor
Hence (E) becomes
[tex](x-1)(121x^3-4699x^2+23947x-21417)=0[/tex]
[tex](121x^3-4699x^2+23947x-21417)=\left(x-33\right)\left(121x^2-706x+649\right)[/tex]
(E)= becomes
[tex]\left(x-1\right)\left(x-33\right)\left(121x^2-706x+649\right)=0[/tex]
hence x =1 or x=33
the third factor and not be factorized
Checking our solution for x=1
[tex]\sqrt{3\cdot \:1-1}+\sqrt{5\cdot \:1-3}+\sqrt{1-1}=2\sqrt{2}[/tex]
[tex]2\sqrt{2}=2\sqrt{2}[/tex]
hence true
Checking our solution for x=33
[tex]\sqrt{3\cdot \:33-1}+\sqrt{5\cdot \:33-3}+\sqrt{33-1}=2\sqrt{2}[/tex]
[tex]20\sqrt{2}=2\sqrt{2}[/tex]
False
hence our answer is
x=1
