Question

ASAP! I really need help with this question! No nonsense answers, and please attach the solution.

Answer 1

Answer:

$\boxed{\sf Option \ 4}$

Step-by-step explanation:

$\sqrt{2x-3} +x=3$

Subtract x from both sides.

$\sqrt{2x-3} +x-x=-x+3$

$\sqrt{2x-3}=-x+3$

Square both sides.

$( \sqrt{2x-3})^2 =(-x+3)^2$

$2x-3=x^2-6x+9$

Subtract x²-6x+9 from both sides.

$2x-3-(x^2-6x+9 )=x^2-6x+9-(x^2-6x+9)$

$-x^2 +8x-12=0$

Factor left side of the equation.

$(-x+2)(x-6)=0$

Set factors equal to 0.

$-x+2=0\\-x=-2\\x=2$

$x-6=0\\x=6$

Check if the solutions are extraneous or not.

Plug x as 2.

$\sqrt{2(2)-3} +2=3\\ \sqrt{4-3} +2=3\\\sqrt{1} +2=3\\3=3$

x = 2 works in the equation.

Plug x as 6.

$\sqrt{2(6)-3} +6=3\\ \sqrt{12-3} +6=3\\\sqrt{9} +6=3\\3+6=3\\9=3$

x = 6 does not work in the equation.

Answer 2

Answer:

option d

Step-by-step explanation:

$\sqrt{2x-3}+x = 3\\\\\sqrt{2x-3} = 3 -x$

Square both sides

$(\sqrt{2x-3})^{2}=(3-x)^{2}\\\\\\2x-3=9-6x+x^{2}\\\\0=x^{2}-6x + 9 - 2x + 3$   {Add like terms}

$x^{2} - 8x + 12 = 0$

Sum = -8

Product = 12

Factors =  -2 , - 6

x² - 2x - 6x + (-2) * (-6) = 0

x(x -2) - 6(x -2) = 0

(x -2) (x - 6) = 0

x - 2 =0   ; x - 6 = 0

x = 2      ; x = 6

roots of the equation : 2  , 6

But when we put x = 6, it doesn't satisfies the equation.

When x = 6,

$\sqrt{2x-3} + x = 3\\\\\sqrt{2*6-3}+6 = 3\\\\\sqrt{12-3}+6=3\\\\\sqrt{9}+6=3$

3 + 6≠ 3

Therefore, x = 2  but x = 6 is extraneous