How do you solve #5^3 sqrt(4x - 2) = 17 #?

3 Answers
EZ as pi
Sep 30, 2017

#x = 0.504624#

Explanation:

Square both sides to get rid of the square root.

#(5^3sqrt(4x-2))^2 = 17^2#

#5^6 xx (4x-2) = 289#

#15625 (4x-2) = 289" "larr# isolate the bracket with #x#

#color(white)(xxxxxxxxxxxxxxxxxxxx)#Divide both sides by #15625#

#4x-2 = 289/15625" "larr# isolate the term in #x#

#4x = 289/15625 +2#

#4x = 2.018496" "larr div 4# to isolate #x#

#x = 2.018496/4#

#x = 0.504624#

George C.
Sep 30, 2017

If the intended equation was:

#5 root(3)(4x-2) = 17#

then #x = 5163/500 = 10.326#

Explanation:

Suppose the intended equation was:

#5 root(3)(4x-2) = 17#

We can cube both sides to get:

#125(4x-2) = 4913#

which multiplies out to get:

#500x-250 = 4913#

Add #250# to both sides to get:

#500x = 5163#

Divide both sides by #500# to get:

#x = 5163/500 = 10.326#

Barney V.
Sep 30, 2017

#color(magenta)(x=0.504624#

Explanation:

#5^3sqrt(4x-2)=17#

#:.sqrt(4x-2)=17/125#

Square both sides

#:.(sqrt(4x-2))^2=(17/125)^2#

#:.(sqrt(4x-2))^2=(0.136)^2#

#:.sqrtaxxsqrta=a#

#:.4x-2=0.018496#

#:.4x=2.018496#

#:.x=2.018496/4#

#:.color(magenta)(x=0.504624#

~~~~~~~~~~~~~~~~~

check:-

substitute #color(magenta)(x=0.504624#

#:.5^3sqrt(4(color(magenta)0.504624)-2)=17#

#:.125sqrt(0.018496)=17#

#:.125xx0.136=17#

#:.color(magenta)(17=17#

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