Question

WILL GIVE BRAINIEST TO THE FIRST ANSWER NEED HELP ASAP

Answer 1

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