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solve `x+sqrt(2x-1)=8`  

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user5451174 | eNotes Newbie

Posted June 26, 2013 at 9:04 AM via web

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

 

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Mary Joy Ripalda | High School Teacher | (Level 3) Educator

Posted June 26, 2013 at 10:37 AM (Answer #3)

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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.

`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.

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Mary Joy Ripalda | High School Teacher | (Level 3) Educator

Posted June 26, 2013 at 12:16 PM (Reply #1)

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

 

But when we plug-in values of the varaible 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 (+).    

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llltkl | College Teacher | Valedictorian

Posted June 26, 2013 at 1:45 PM (Reply #2)

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Yes! Thank you everybody for your responses and valuable inputs.

In strict mathematical terms both 5 and 13 are solutions to this radical equation. But x = 13 constitutes an extraneous solution because the principal square root of 25 is 5 which needs to be considered in this case. Therefore, the solution of x that satisfies above radical equation is x=5 only.

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aruv | High School Teacher | Valedictorian

Posted June 26, 2013 at 10:20 AM (Answer #2)

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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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aruv | High School Teacher | Valedictorian

Posted June 27, 2013 at 12:40 PM (Answer #5)

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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.

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