How do you solve #sqrt(2x-9)=11# and identify any restrictions?

1 Answer
SierraX · salamat
Mar 24, 2017

#x =65#, restrictions are that #x>=4.5#

Explanation:

#sqrt (2 x -9) = 11#

square both sides

#(sqrt (2 x -9))^2 = 11^2#
#2 x -9 = 121#
#2 x = 121 + 9 =130#
#x =130/2 = 65#

#sqrt (2 x -9)# should be greater than or equal to zero, since we can't take the square roots of negative values. Restrictions can be found by designating the radicand (the stuff under the square root) as equal to zero.

#2x-9=0#
#2x=9#
#x=4.5#

An x value of 4.5 will make the radicand 0, which is the minimum number it can be.

Therefore, the restrictions for x are any numbers under 4.5.

Related questions
See all questions in Radical Equations
Impact of this question
1671 views around the world
You can reuse this answer
Creative Commons License