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mel-nik [20]
3 years ago
7

WILL GIVE BRAINIEST TO THE FIRST ANSWER NEED HELP ASAP

Mathematics
1 answer:
cestrela7 [59]3 years ago
5 0

Answer:

x=1

Step-by-step explanation:

\sqrt{3x-1}+\sqrt{5x-3}+\sqrt{x-1}=2\sqrt{2}

\sqrt{3x-1}+\sqrt{5x-3}+\sqrt{x-1}-\sqrt{x-1}=2\sqrt{2}-\sqrt{x-1}

\sqrt{3x-1}+\sqrt{5x-3}=2\sqrt{2}-\sqrt{x-1}

Squaring on both sides we get

\left(\sqrt{3x-1}+\sqrt{5x-3}\right)^2=\left(2\sqrt{2}-\sqrt{x-1}\right)^2 -----(A)

Expanding Left hand side we get

\left(\sqrt{3x-1}+\sqrt{5x-3}\right)^2=\left(2\sqrt{2}-\sqrt{x-1}\right)^2

 =\left(\sqrt{3x-1}\right)^2+2\sqrt{3x-1}\sqrt{5x-3}+\left(\sqrt{5x-3}\right)^2

 =\left(3x-1\right)+2\sqrt{3x-1}\sqrt{5x-3}+\left(5x-3\right)

 =8x+2\sqrt{3x-1}\sqrt{5x-3}-4

Expanding Right Hand side

\left(2\sqrt{2}-\sqrt{x-1}\right)^2

  = \left(2\sqrt{2}\right)^2-2\cdot \:2\sqrt{2}\sqrt{x-1}+\left(\sqrt{x-1}\right)^2

 = 8-4\sqrt{2}\sqrt{x-1}+\left(x-1\right)

 = x+7-4\sqrt{2}\sqrt{x-1}

Hence we have now (A) equal to

8x+2\sqrt{3x-1}\sqrt{5x-3}-4=x+7-4\sqrt{2}\sqrt{x-1}

subtracting 8x from both sides

8x+2\sqrt{3x-1}\sqrt{5x-3}-4-8x=x+7-4\sqrt{2}\sqrt{x-1}-8x

2\sqrt{3x-1}\sqrt{5x-3}-4=-7x-4\sqrt{2}\sqrt{x-1}+7

Adding 4 on both sides we get...

2\sqrt{3x-1}\sqrt{5x-3}-4+4=-7x-4\sqrt{2}\sqrt{x-1}+7+4

2\sqrt{3x-1}\sqrt{5x-3}=-7x+11-4\sqrt{2}\sqrt{x-1}

now squaring on both sides again

\left(2\sqrt{3x-1}\sqrt{5x-3}\right)^2=\left(-7x+11-4\sqrt{2}\sqrt{x-1}\right)^2

                                                                   ----------(B)

Left hand side:

\left(2\sqrt{3x-1}\sqrt{5x-3}\right)^2

  = 4\left(3x-1\right)\left(5x-3\right)

  = 60x^2-56x+12

Right hand side:

\left(-7x+11-4\sqrt{2}\sqrt{x-1}\right)^2

  = \left(-7x+11-4\sqrt{2}\sqrt{x-1}\right)\left(-7x+11-4\sqrt{2}\sqrt{x-1}\right)

  = 49x^2-154x+56\sqrt{2}\sqrt{x-1}x+121-88\sqrt{2}\sqrt{x-1}+32\left(x-1\right)

  = 49x^2+56\sqrt{2}x\sqrt{x-1}-122x-88\sqrt{2}\sqrt{x-1}+89

Hence we have (B) equals to

60x^2-56x+12=49x^2+56\sqrt{2}x\sqrt{x-1}-122x-88\sqrt{2}\sqrt{x-1}+89

49x^2+56\sqrt{2}x\sqrt{x-1}-122x-88\sqrt{2}\sqrt{x-1}+89=60x^2-56x+12

Subtract 49x^2-122x from both sides

56\sqrt{2}x\sqrt{x-1}-88\sqrt{2}\sqrt{x-1}+89=11x^2+66x+12

56\sqrt{2}x\sqrt{x-1}-88\sqrt{2}\sqrt{x-1}=11x^2+66x-77  ---(C)

Also

56\sqrt{2}x\sqrt{x-1}-88\sqrt{2}\sqrt{x-1}

 = 7\cdot \:8\sqrt{x-1}\sqrt{2}x-11\cdot \:8\sqrt{x-1}\sqrt{2}

 = 8\sqrt{x-1}\sqrt{2}\left(7x-11\right)

Hence (C) becomes

8\sqrt{2}\sqrt{x-1}\left(7x-11\right)=11x^2+66x-77

Squaring both sides

\left(8\sqrt{2}\sqrt{x-1}\left(7x-11\right)\right)^2=\left(11x^2+66x-77\right)^2

                                      -------(D)

Left Hand Side

= 6272\left(x-1\right)x^2-19712\left(x-1\right)x+15488\left(x-1\right)

=  6272x^3-25984x^2+35200x-15488

Right hand Side

= \left(11x^2+66x-77\right)\left(11x^2+66x-77\right)

=  121x^4+1452x^3+2662x^2-10164x+5929

Hence (D) becomes

6272x^3-25984x^2+35200x-15488=121x^4+1452x^3+2662x^2-10164x+5929

121x^4+1452x^3+2662x^2-10164x+5929=6272x^3-25984x^2+35200x-15488

subtracting 6272x^3-25984x^2+35200x-15488 from both sides

121x^4-4820x^3+28646x^2-45364x+21417=0 -----(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

(x-1)(121x^3-4699x^2+23947x-21417)=0

(121x^3-4699x^2+23947x-21417)=\left(x-33\right)\left(121x^2-706x+649\right)

(E)= becomes

\left(x-1\right)\left(x-33\right)\left(121x^2-706x+649\right)=0

hence x =1 or x=33

the third factor and not be factorized

Checking our solution for x=1

\sqrt{3\cdot \:1-1}+\sqrt{5\cdot \:1-3}+\sqrt{1-1}=2\sqrt{2}

2\sqrt{2}=2\sqrt{2}

hence true

Checking our solution for x=33

\sqrt{3\cdot \:33-1}+\sqrt{5\cdot \:33-3}+\sqrt{33-1}=2\sqrt{2}

20\sqrt{2}=2\sqrt{2}

False

hence our answer is

x=1

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