1 Answer

Tomas By · Answer 1 · 2023-03-16T20:54:37+0000

Given the following equation:

We will solve the equation as follows:

First, square both sides to remove the square root

$\begin{gathered} (√(22-7x))^2=(x-4)^2 \\ 22-7x=x^2-8x+16 \end{gathered}$

Combine the like terms:

Factor the equation:

$\begin{gathered} (x+2)(x-3)=0 \\ x+2=0\rightarrow x=-2 \\ x-3=0\rightarrow x=3 \end{gathered}$

Now, we will check x = -2

$\begin{gathered} x=-2\rightarrow√(22-7(-2))=√(36)=6 \\ x=-2\operatorname{\rightarrow}x-4=-2-4=-6 \\ 6\\e-6 \end{gathered}$

So, x = -2 is not a valid solution

Check the value of x = 3

$\begin{gathered} x=3\operatorname{\rightarrow}√(22-7(3))=√(22-21)=1 \\ x=3\operatorname{\rightarrow}x-4=3-4=-1 \\ 1\\e-1 \end{gathered}$

So, x = 3 is not a valid solution

So, the answer will be:

$x=\phi$

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