Question

Solve for all possible values of x.V22 - 7x = 2 -4

Answer 1

Given the following equation:

$\sqrt{22-7x}=x-4$

We will solve the equation as follows:

First, square both sides to remove the square root

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

Combine the like terms:

$x^2-x-6=0$

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

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

So, x = -2 is not a valid solution

Check the value of x = 3

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

So, x = 3 is not a valid solution

So, the answer will be:

$x=\phi$