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You are correct so far. The values of x are 5 and 13.
But take note that in radical equations, after solving for the values of x, it is necessary to plug-in these values to the original equation. Since there are instances that the solved values don't work on the original equation.
So let's plug-in x=5 to check.
Since the resulting condition is True, then x=5 is a solution to the equation.
Next, plug-in x=13 to check too.
`18=8 ` (False)
Notice that left side does not simplify to 8. So, x=13 is not a solution to the given equation.
Therefore, the solution to `x+sqrt(2x-1)=8` is `x=5` only.
X=13 will not satisfy the given equation so it can not be root of the equation.
We have equation
squaring both side
But x=13 is not possible because
put x=13 in the given equation
This root is known as extraneous root of the equation.
Thus solution of problem is x=5.
Let define a function f(x) as
`x+sqrt(2x-1)=8` ,it is given equation.
The zero of f(x) is the root of the equation `x+sqrt(2x-1)=8`
Now see above graph of f(x) , it meet x-axis only at x=5.
To say that 5 is zero of f(x) .
So 5 is root of the equation `x+sqrt(2x-1)=8`
So now it is clear from this approach that that we can solve radical equation graphically as well.It has only one solution.
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