How do you solve #x-4= sqrt(2x) # and find any extraneous solutions?

1 Answer
Dec 25, 2016

Answer:

#x=8#

Explanation:

Before solving the above equation, you must have some background knowledge.

  • The above equation is quadratic equation.
  • Quadratic equations are of form #ax^2+bx+c=0# where a, b and c are real numbers.
  • It has two roots which means it has two solutions.
  • Its possible that one solution may repeat.
  • Some quadratic equations can be solved by factorization.
  • The solutions of some quadratic equations are not rational and cannot be factorized.
    For such equations, the most common method of solution is the quadratic formula.
    The quadratic formula can be used to solve ANY quadratic equation, even those which cannot be factorized.
  • Quadratic Formula: #x=(-b+-sqrt(b^2-4ac))/(2a)#
  • The quantity #b^2-4ac# in the quadratic formula is called discriminant.

If #b^2-4ac##>##0#, then there are two real and distinct solutions to the quadratic equation.

If #b^2-4ac##=##0#, then there are two equal solutions to the quadratic equation.

If #b^2-4ac##<##0#, then there are two complex solutions to the quadratic equation.

Now, we have to convert #x-4=sqrt(2x)# in quadratic form i.e

#ax^2+bx+c=0#

Squaring both sides of #x-4=sqrt(2x)# we have,

#(x-4)^2=(sqrt(2x))^2#

or, #x^2-2*x*4+4^2=2x rarr# Using #(a-b)^2=a^2-2ab+b^2#

formula where #a=x# and #b=4#

or, #x^2-8x+16=2x#

or, #x^2-8x-2x+16=0#

or, #x^2-10x+16=0#

or, #1x^2+(-10)x+16=0rarr#Now it is in quaratic form where #a=1#, #b=-10# and #c=16#

Now using the quadratic formula, we have

#x=(-b+sqrt(b^2-4ac))/(2a)rarr#First use + sign

or, #x=(-(-10)+sqrt((-10)^2-4*1*16))/(2*1)#

or, #x=(10+sqrt(100-64))/2#

or, #x=(10+sqrt(36))/2#

or, #x=(10+6)/2#

or, #x=16/2#

or, #x=8# First root (solution)

Again, we have to simplify using - sign in formula to get second root.

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

or, #x=(-(-10)-sqrt((-10)^2-4*1*16))/(2)#

or, #x=(10-sqrt(100-64))/(2)#

or, #x=(10-sqrt(36))/(2)#

or, #x=(10-6)/(2)#

or, #x=4/2=2rarr#Second root (solution)

When #x=8#

Then #x-4=sqrt(2x)#

or, #8-4==sqrt(2*8)#

or, #4=sqrt(16)#

or, #4=4rarr#They check.

When #x=2#

Then #x-4=sqrt(2x)#

or, #2-4=sqrt(2*2)#

or, #-2=sqrt(4)#

or, #-2cancel(=)2rarr#They don't check.

So, the only value of #x# is #8#.

Lastly, I have made it this long so that you won't encounter any problems with quadratic equation in future.