# Please solve and show your work. Solve (x-4)^2 = 3.

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You may use the alternate method such that:

` (x-4)^2 = 3 => (x-4)^2 - 3 = 0 => (x-4)^2 - (sqrt3)^2 = 0`

You need to convert the difference of squares into a product using the following formula such that:

`a^2 - b^2 = (a - b)(a + b)`

Reasoning by analogy yields:

`(x-4)^2 - (sqrt3)^2 = (x - 4 - sqrt3)(x - 4 + sqrt3)`

Since `(x-4)^2 - (sqrt3)^2 = 0` , then `(x - 4 - sqrt3)(x - 4 + sqrt3) ` `= 0` , hence, you need to solve the following equations such that:

`{(x - 4 - sqrt3 = 0),(x - 4+ sqrt3 = 0):}=> {(x= 4+ sqrt3),(x= 4- sqrt3):}`

**Hence, evaluating the solutions to the given equation yields `x = 4 +sqrt3` and `x = 4 - sqrt3.` **

The equation (x-4)^2 = 3 has to be solved.

Take the square roots of both the sides.

`sqrt(x-4)^2 = sqrt (3)`

Now, the square of both +- x is equal to x^2. This givesL

`x - 4 = +-sqrt 3`

x = `+sqrt + 4` and x = `-sqrt 3 + 4`

The solutions of the equation (x-4)^2 = 3 are `(sqrt 3 + 4, -sqrt 3 + 4)`

Another simpler method to solve (x-4)^2 = 3 is as under:

Take sqrt of both sides:

sqrt[(x-4)^2] = sqrt(3)

=> x-4 = +sqrt(3), -sqrt(3) as any number have positive and negative roots

Add 4 to both sides:

x-4+4 = sqrt(3)+4 , -sqrt(3)+4

x = 4+sqrt(3) , 4-sqrt(3)

**Hence the solution of (x-4)^2=3 is x = 4+sqrt(3) , 4-sqrt(3)**

We have to solve (x-4)^2 = 3

=> x^2-8x+16 = 3

=> x^2-8x+13 = 0

This is in the form ax^2+bx+c=0 where a=1, b=-8 and c=13 and its roots are as:

x= [-b+sqrt(b^2-4ac)]/2a , [-b-sqrt(b^2-4ac)]/2a

Hence the roots are:

x = [-(-8)+sqrt{(-8)^2-4*1*13}]/(2*1) and

x = [-(-8)-sqrt{(-8)^2-4*1*13}]/(2*1)

=> x = [8+sqrt{64-52}]/2 , [8-sqrt{64-52}]/2

=> x = [8+sqrt{12}]/2 , [8-sqrt{12}]/2

=> x = [8+2sqrt(3)]/2 , [8-2sqrt(3)]/2

=> x = 4+sqrt(3) , 4-sqrt(3)

**The solution of equation (x-4)^2 = 3 **

**is x=4+sqrt(3),4-sqrt(3)**