solve `x+sqrt(2x-1)=8`  

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lemjay's profile pic

lemjay | High School Teacher | (Level 3) Senior Educator

Posted on

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.






`8=8 `    

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.

aruv's profile pic

aruv | High School Teacher | (Level 2) Valedictorian

Posted on

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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aruv | High School Teacher | (Level 2) Valedictorian

Posted on

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.

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