How do you solve \sqrt{x}=x-6x=x6?

3 Answers
May 2, 2018

x = 9x=9

Explanation:

sqrt(x) = x- 6x=x6
Square the equation:
x = (x-6)^2x=(x6)2
Apply the expansion of (a- b)^2 = a^2 -2ab + b^2(ab)2=a22ab+b2
implies x = x^2 - 12x + 36x=x212x+36
implies 0 = x^2 - 13x + 360=x213x+36
Factorize the quadratic.
implies x^2 - 9x -4x + 36 = 0x29x4x+36=0
implies x(x-9)-4(x-9) = 0x(x9)4(x9)=0
implies (x-4)(x-9) = 0(x4)(x9)=0
implies x = 4 or x = 9x=4orx=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

x = 9x=9

Explanation:

sqrtx = x - 6x=x6

First, square both sides:
sqrtx^color(red)(2) = (x-6)^color(red)2x2=(x6)2

Simplify:
x = x^2 - 12x + 36x=x212x+36

Move everything to one side of the equation:
0 = x^2 - 13x + 360=x213x+36

Now we need to factor.
Our equation is standard form, or ax^2 + bx + cax2+bx+c.

The factored form is (x-m)(x-n)(xm)(xn), where mm and nn are integers.

We have two rules to find mm and nn:

  • mm and nn have to multiply up to a * cac, or 3636
  • mm and nn have to add up to bb, or -1313

Those two numbers are -44 and -99. So we put them into our factored form:

0 = (x-4)(x-9)0=(x4)(x9)

Therefore,

x - 4 = 0x4=0 and x - 9 = 0x9=0

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

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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

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.