Respuesta :
The given condition may be mathematically expressed as,
sqrt (x) + 9 - 4 = 1
Group all the constants to the same side of the equation,
sqrt (x) = -4
Squaring both sides give an answer of x = 16.
The answer for the question above is x = 16 is not an extraneous root because if we substitute this value to the x of the equation, the equation becomes true.
sqrt (x) + 9 - 4 = 1
Group all the constants to the same side of the equation,
sqrt (x) = -4
Squaring both sides give an answer of x = 16.
The answer for the question above is x = 16 is not an extraneous root because if we substitute this value to the x of the equation, the equation becomes true.
Remember that an extraneous solution of an equation is a solution that emerges from the algebraic process of solving the equation but is not a valid solution of the equation.
First, we are going to solve our equation algebraically:
Step 1 simplify the equation:
[tex] \sqrt{x} +9-4=1 [/tex]
[tex] \sqrt{x}+ 5=1 [/tex]
Step 2 subtract 5 from both sides of the equation:
[tex] \sqrt{x} +5-5=1-5 [/tex]
[tex] \sqrt{x} =-4 [/tex]
Step 3 square both sides of the equation:
[tex] \sqrt{x} ^2=(-4)^2 [/tex]
[tex] x=16 [/tex]
Next, we are going to replace our solution in our original equation and check if it is a valid solution:
[tex] \sqrt{x} +9-4=1 [/tex]
[tex] \sqrt{16} +5=1 [/tex]
[tex] 4+5=1 [/tex]
[tex] 9\neq 1 [/tex]
Since 9 is not equal to 1, [tex] x=16 [/tex] is not valid solution of the equation; therefor it is an extraneous solution.
We can conclude that the correct answer is: x = 16, solution is extraneous