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lemjay eNotes educator| Certified Educator

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.

lemjay eNotes educator| Certified Educator

Note that we only apply the square root property if we are solving for the value of a variable in an equation.

But when we plug in values of the variable to the radical equation, we do not apply the square root property.

So, when we take the square root of 25, it would be positive 5, since the sign outside `sqrt(2x-1)` is plus (+).

aruv | Student

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.

aruv | Student

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.

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