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malfutka [58]
3 years ago
13

Solve for x, given the equation Square root of x plus 9 − 4 = 1.

Mathematics
2 answers:
finlep [7]3 years ago
8 0

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:

\sqrt{x} +9-4=1

\sqrt{x}+ 5=1

Step 2 subtract 5 from both sides of the equation:

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

\sqrt{x} =-4

Step 3 square both sides of the equation:

\sqrt{x} ^2=(-4)^2

x=16

Next, we are going to replace our solution in our original equation and check if it is a valid solution:

\sqrt{x} +9-4=1

\sqrt{16} +5=1

4+5=1

9\neq 1

Since 9 is not equal to 1, x=16 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

Len [333]3 years ago
5 0
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. 
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