How do you solve #x= sqrt(6x) +7#?

2 Answers
Apr 15, 2018

Answer:

#x=10+sqrt51#

Explanation:

#x=sqrt(6x)+7#
First let's define the domain :
#x>=0#
So:
#x-7=sqrt(6x)#
#(x-7)²=6x#
#(x²-14x+49)-6x=0#
#x²-20x+49=0#
#Δ=20²-4*49#
#Δ=204#
#x=(-b+-sqrtΔ)/2a#
#x_"1"=(20-sqrt204)/2#
#x_"2"=(20+sqrt204)/2#
#x_"1"=(20-2sqrt51)/2#
#x_"2"=(20+2sqrt51)/2#
#x_"1"=10-sqrt51<0=># NOT a solution.
#x_"2"=10+sqrt51#
So : #x=10+sqrt51#
\0/ herex's our answer!

Apr 15, 2018

Answer:

#x=10+sqrt51#

Bare with me please: this is a long solution process:

Explanation:

We can subtract #7# from both sides to arrive at:

#x-7=sqrt(6x)#

We can square both sides to get:

#color(blue)((x-7)^2)=6x#

#=color(blue)((x-7)(x-7))=6x#

What I have in blue, we can use the highly useful mnemonic FOIL to simplify this. We simply multiply the first terms, outside terms, inside and last terms. We get:

  • First terms: #x*x=x^2#
  • Outside terms: #x*-7=-7x#
  • Inside terms: #-7*x=-7x#
  • Last terms: #-7*-7=49#

Now, we have:

#color(blue)(x^2-7x-7x+49)=6x#

Which can be simplified to

#x^2-14x+49=6x#

We have a quadratic on the left, so we want to set it equal to zero to find its zeroes. We do this by subtracting #6x# from both sides to get:

#x^2-20x+49=0#

The only factors of #49# are #1, 7# and #49#, and neither of them, positive or negative, add up to #-20#. We can use the quadratic formula:

#x=(-b+-sqrt(b^2-4ac))/(2a)#

where #a=1, b=-20 & c=49# (in the form #ax^2+bx+c=0#)

Plugging into the quadratic formula, we get:

#x=(20+-sqrt(400-4(1*49)))/(2*1)#

Which simplifies to

#x=(20+-color(blue)(sqrt(204)))/2#

Because #51*4=204#, we can rewrite #color(blue)(sqrt204)# as

#color(blue)(sqrt4*sqrt51)#

Which obviously simplifies to #color(blue)(2sqrt51)#

Now, we have

#x=(20+-2sqrt51)/2#

Every term is divisible by #2#, so we can factor a 2 out to get

#x=(color(red)(10)cancel(20)+-cancel2sqrt51)/(color(red)(1)cancel2)#

Which is equal to

#color(red)(x=10+-sqrt51)#

Since our answer had a square root in it, we know the domain has to be #x>=0#.

#10+sqrt51>0#, so this is a solution for #x# but

#10-sqrt51<0#, which is outside of the domain, so we get

#x=10+sqrt51#

as our final solution.

This was a long solution process, but all I did was:

  • Get #x# on one side
  • Square both sides to get rid of the square root
  • Used FOIL to simplify the left side
  • Used the Quadratic Formula
  • Checked the domain

I really hope this helps!