Question

How do you solve `\sqrt { 169- x ^ { 2} } + 17= x`?

Answer 1

Null set

Explanation:

`sqrt(169-x^2)+17=x`

`sqrt(169-x^2)=x-17`

`169-x^2=(x-17)^2`

`169-x^2=x^2-34x+289`

`2x^2+34x-120=0`

`x^2+17x-60=0`

`(x+20)*(x-3)=0`

But, neither `x=-20` nor `x=3` are solutions of this equation. Hence, solution of it is null set.

Answer 2

`color(blue)("No solutions for " x in RR)`

Explanation:

`sqrt(169-x^2)+17=x`

Subtract 17 from both sides:

`sqrt(169-x^2)=x-17`

Square both sides:

`169-x^2=(x-17)^2`

Expand RHS:

`169-x^2=x^2-34x+289`

Simplify:

`2x^2-34x+120=0`

Factor:

`(2x-10)(x-12)=0=>x=5 and x=12`

Checking solutions:

`x=5`

`sqrt(169-(5)^2)+17=5`

`12+17=5color(white)(88)` False

`-12+17=5color(white)(88)` True

`x=12`

`sqrt(169-(12)^2)+17=12`

`5+17=12color(white)(88)` False

`-5+17+12color(white)(88)`True

We can see that the roots found are only solutions if we take the negative root of `sqrt(169-x^2)`, since we assume `sqrt(169-x^2)` to be the positive root no solutions exist:

Answer 3

Solution: `x = 12 , x =5`

Explanation:

`sqrt( 169- x^2)+17 = x` or

`x -17 = sqrt( 169- x^2)` squaring both sides we get,

`(x -17)^2 = 169- x^2` or

`x^2 -34 x +289= 169- x^2` or

`2 x^2 -34 x +289- 169 = 0` or

`2 x^2 -34 x +120 = 0` or

`2 (x^2 -17 x +60) = 0` or

`x^2 -17 x +60 = 0` or

`x^2 -12 x -5 x +60 = 0` or

`x(x -12) -5(x -12) = 0` or

`(x -12)( x -5) = 0 :. x = 12 , x =5`

Solution: `x = 12 , x =5` [Ans]