How do you solve #\sqrt{x}=x-6#?

3 Answers
May 2, 2018

Answer:

#x = 9#

Explanation:

#sqrt(x) = x- 6#
Square the equation:
#x = (x-6)^2#
Apply the expansion of #(a- b)^2 = a^2 -2ab + b^2#
#implies x = x^2 - 12x + 36#
#implies 0 = x^2 - 13x + 36#
Factorize the quadratic.
#implies x^2 - 9x -4x + 36 = 0#
#implies x(x-9)-4(x-9) = 0#
#implies (x-4)(x-9) = 0#
#implies x = 4 or x = 9#

Note that substituting 4 in the equation returns 2 = -2, which is obviously wrong. So we neglect x = 4 in the set of solutions. Take care to verify your answers after solving(don't make my mistake!)

May 2, 2018

Answer:

#x = 9#

Explanation:

#sqrtx = x - 6#

First, square both sides:
#sqrtx^color(red)(2) = (x-6)^color(red)2#

Simplify:
#x = x^2 - 12x + 36#

Move everything to one side of the equation:
#0 = x^2 - 13x + 36#

Now we need to factor.
Our equation is standard form, or #ax^2 + bx + c#.

The factored form is #(x-m)(x-n)#, where #m# and #n# are integers.

We have two rules to find #m# and #n#:

  • #m# and #n# have to multiply up to #a * c#, or #36#
  • #m# and #n# have to add up to #b#, or #-13#

Those two numbers are #-4# and #-9#. So we put them into our factored form:

#0 = (x-4)(x-9)#

Therefore,

#x - 4 = 0# and #x - 9 = 0#

#x = 4# #quadquadquad# and #quadquadquad# ## #x = 9#

#--------------------#

However, we still need to check our answers by substituting them back into the original equation, since we have a square root in our original equation.

Let's first check if #x = 4# is really a solution:
#sqrt4 = 4 - 6#

#2 = -2#

This is not true! That means that #x !=4# (#4# is not a solution)

Now let's check #x = 9#:
#sqrt9 = 9 - 6#

#3 = 3#

This is true! That means that #x = 9# (#9# is really a solution)

So the final answer is #x = 9#.

Hope this helps!

May 2, 2018

Answer:

#x=9# is the only real solution to this equation.

Explanation:

First, square both sides of this equation.

#x=x^2-12x+36#

Now put in standard form.

#x^2-13x+36=0#

Factor.

#(x-4)(x-9)=0#

#x=9# is a solution to this equation. #x=4# is not a solution to the original equation. However it is a solution to

#x=x^2-12x+36#

When we squared both sides to at the beginning, we enabled an extraneous solution since #(-sqrtx)^2=(sqrtx)^2=x#. Thus we enabled #-sqrtx# as a valid left-hand side of the equation when the original problem did not. Note that #-sqrtx=x-6# when #x=4#, but this is not what the problem is asking.