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

3 Answers
Cem Sentin
Jul 3, 2018

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.

Somebody N.
Jul 3, 2018

#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:

Binayaka C.
Jul 3, 2018

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]

Related questions
Impact of this question
2683 views around the world
You can reuse this answer
Creative Commons License