# solve `x+sqrt(2x-1)=8`

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.

`x+sqrt(2x-1)=8`

`5+sqrt(2*5-1)=8`

`5+sqrt(10-1)=8`

`5+sqrt9=8`

`5+3=8`

`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.

`13+sqrt(2*13-)=8`

`13+sqrt(26-1)=8`

`13+sqrt25=8`

`13+5=8`

`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.

Posted on

Let define a function f(x) as

`f(x)=x+sqrt(2x-1)-8`

Note:

`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.

Posted on

X=13 will not satisfy the given equation so it can not be root of the equation.

We have equation

`x+sqrt(2x-1)=8`

`sqrt(2x-1)=8-x`

squaring both side

`2x-1=(8-x)^2`

`2x-1=64+x^2-16x`

`x^2-16x-2x+1+64=0`

`x^2-18x+65=0`

`x^2-13x-5x+65=0`

`x(x-13)-5(x-13)=0`

`(x-5)(x-13)=0`

`either`

`x-5=0`

`x=5`

`or`

`x-13=0`

`x=13`

But x=13 is not possible because

put x=13 in the given equation

`13+sqrt(2xx13-1)=13+sqrt(26-1)=13+5=18!=5`

This root is known as extraneous root of the equation.

Thus solution of problem is x=5.