How do you solve #\root[ 4] { x ^ { 2} + 6x } - 2= 0#?

1 Answer
Aug 2, 2017

#x = -8, 2#

Explanation:

We have: #root(4)(x^(2) + 6 x) - 2 = 0#

#Rightarrow (x^(2) + 6 x)^(frac(1)(4)) - 2 = 0#

First, let's add #2# to both sides of the equation:

#Rightarrow (x^(2) + 6 x)^(frac(1)(4)) = 2#

Then, let's raise both sides to #4#:

#Rightarrow ((x^(2) + 6 x)^(frac(1)(4)))^(4) = (2)^(4)#

#Rightarrow x^(2) + 6 x = 16#

Now, let's subtract both sides by #16#, and then use the quadratic formula:

#Rightarrow x^(2) + 6 x - 16 = 0#

#Rightarrow x = frac(- 6 pm sqrt((6)^(2) - 4(1)(- 16))(2(1))#

#Rightarrow x = frac(- 6 pm sqrt(36 + 64))(2)#

#Rightarrow x = frac(- 6 pm sqrt(100))(2)#

#Rightarrow x = frac(- 6 pm 10)(2)#

#Rightarrow x = - 8, 2#

Finally, let's verify these solutions by substituting them back into the original equation:

#Rightarrow root(4)((- 8)^(2) + 6 cdot (- 8)) - 2 = 0#

#Rightarrow root(4)(64 - 48) - 2 = 0#

#Rightarrow root(4)(16) - 2 = 0#

#Rightarrow 2 - 2 = 0#

#Rightarrow 0 = 0 " " "True"#

#and#

#Rightarrow root(4)((2)^(2) + 6 cdot (2)) - 2 = 0#

#Rightarrow root(4)(4 + 12) - 2 = 0#

#Rightarrow root(4)(16) - 2 = 0#

#Rightarrow 2 - 2 = 0#

#Rightarrow 0 = 0 " " "True"#

Therefore, the solutions to the equation are #x = - 8# and #x = 2#.