How do you solve #j^ { 2} - 24j - 29= 0#?

1 Answer
George C.
Nov 19, 2016

#j = 12 +-sqrt(173)#

Explanation:

Complete the square, then use the difference of squares identity:

#a^2-b^2 = (a-b)(a+b)#

with #a=(j-12)# and #b=sqrt(173)# as follows:

#0 = j^2-24j-29#

#color(white)(0) = j^2-24j+144-173#

#color(white)(0) = (j-12)^2-(sqrt(173))^2#

#color(white)(0) = ((j-12)-sqrt(173))((j-12)+sqrt(173))#

#color(white)(0) = (j-12-sqrt(173))(j-12+sqrt(173))#

Hence:

#j = 12 +-sqrt(173)#

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