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`sqrt(9y-196)`  + `sqrt(196)` = `sqrt(49)`    Simplify. I do not think there are...

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novjb | Salutatorian

Posted May 26, 2013 at 11:16 PM via web

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`sqrt(9y-196)`  + `sqrt(196)` = `sqrt(49)` 



I do not think there are any solutions, is that right?

Thanks in advance :)

Tagged with algebra, expressions, math, radicals

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jgeertz | College Teacher | (Level 1) Associate Educator

Posted May 27, 2013 at 12:28 AM (Answer #2)

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You are absolutely correct. There are no solutions. Let's solve the problem to see why.

`sqrt(9y-196) +sqrt(196) = sqrt49`

Remove all the perfect squares from under the radical. In this problem both 7 and 14 are perfect squares.

`sqrt(9y-196) +14=7`

Subtract 14 from both sides of the equation.

`sqrt(9y-196) =-7`

Now to eliminate the radical on the left side of the equation, square both sides of the equation.

`(sqrt(9y-196))^2 =(-7)^2`


`9y-196 =49`

Add 196 to both sides of the equation.


Divide both sides by 9 and simplify.

`y= 245/9`

In order to be a valid solution, it must work when substituted into the original equation.

`sqrt(9(245/9)-196) +sqrt196 =sqrt49`

`sqrt(245-196) + 14 =7`

`sqrt49 +14=7`


21 not equal to 7

Therefore the problem has no solution.


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oldnick | Valedictorian

Posted May 27, 2013 at 12:26 AM (Answer #1)

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you have more  equation in one:




`sqrt(9y-196)=+-21`           (1)

`sqrt(9y-196)=+-7`                (2)

  Solving:  (1)

`sqrt(9y-196)= +- 21`


`9y=637`       `rArr y=637/9`

Solving (2):



`9y= 215`       `rArr y=245/9`



 `sqrt(9xx 245/9-196)=` `sqrt(245-196)=sqrt(49)=+-7`


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embizze | High School Teacher | (Level 1) Educator Emeritus

Posted May 27, 2013 at 1:25 AM (Reply #1)

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You are correct in saying that `y^2=196` has two solutions; `y=+-14` . However `sqrt(196)=14` , the principle root. If you want two answers you must have `+-sqrt(196)` .

So the original problem can be rewritten as `sqrt(9y-196)+14=7` or `sqrt(9y-196)=-7` . Thus it is clear there are no solutions as the left hand side is positive while the right hand side is negative.

The graph of the original LHS in black, RHS in red:

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