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Solve the equation and determine the solution(s) or any extraneous solution.

Solve the equation and determine the solution(s) or any extraneous solution.-example-1
User Dambros
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1 Answer

7 votes

Answer:

Solution: x = 9 Extraneous solution: x = 1

Explanation:

Given equation:


√(4x)=x-3

Square both sides:


\implies \left(√(4x)\right)^2=\left(x-3\right)^2


\implies 4x=x^2-6x+9

Subtract 4x from both sides:


\implies 4x-4x=x^2-6x+9-4x


\implies 0=x^2-10x+9


\implies x^2-10x+9=0

Split the term in x:


\implies x^2-x-9x+9=0

Factor the first two terms and the last two terms separately:


\implies x(x-1)-9(x-1)=0

Factor out the common term (x - 1):


\implies (x-9)(x-1)=0

Apply the zero-product property:


(x-9)=0 \implies x = 9


(x-1)=0 \implies x=1

Substitute the found values of x into the original equation to verify the solutions:


\begin{aligned}x=9 \implies √(4 \cdot 9) & = 9-3\\√(36) & = 6\\6 & = 6\end{aligned}


\begin{aligned}x=1 \implies √(4 \cdot 1) & = 1-3\\√(4) & = -2\\ 2 & = -2\end{aligned}

Therefore, the only valid solution is x = 9:

  • Solution: x = 9
  • Extraneous solution: x = 1

Extraneous solution: A solution from the process of solving the equation that is not a valid solution to the original equation.

User Sander Vanhove
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