`sqrt(9y-196)`  + `sqrt(196)` = `sqrt(49)`    Simplify. I do not think there are any solutions, is that right? Thanks in advance :)

2 Answers

jgeertz's profile pic

jgeertz | College Teacher | (Level 1) Associate Educator

Posted on

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.


oldnick's profile pic

oldnick | (Level 1) Valedictorian

Posted on

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's profile pic

embizze | High School Teacher | (Level 2) Educator Emeritus

Posted on

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: