Question

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

Answer 1

`\boxed{\sf Option \ 4}`

Explanation:

`√(2x-3) +x=3`

Subtract x from both sides.

`√(2x-3) +x-x=-x+3`

`√(2x-3)=-x+3`

Square both sides.

`( √(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.

`√(2(2)-3) +2=3\\ √(4-3) +2=3\\√(1) +2=3\\3=3`

x = 2 works in the equation.

Plug x as 6.

`√(2(6)-3) +6=3\\ √(12-3) +6=3\\√(9) +6=3\\3+6=3\\9=3`

x = 6 does not work in the equation.

Answer 2

option d

Explanation:

`√(2x-3)+x = 3\\\\√(2x-3) = 3 -x\\`

Square both sides

`(√(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,

`√(2x-3) + x = 3\\\\√(2*6-3)+6 = 3\\\\√(12-3)+6=3\\\\√(9)+6=3\\\\`

3 + 6≠ 3

Therefore, x = 2 but x = 6 is extraneous