Question

How to solve (x-4)^2-9=0

Answer 1

Explanation:

(x-4)^2-9=0

Let's assume a=x-4 and b=3

We know that,

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

substituting values a, and b in above equation we get,

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

(x-1)(x-7)=0

x=1,7

Answer 2

`\displaystyle{x=7,1}`

Explanation:

Given the equation:

`\displaystyle{\left(x-4\right)^2-9=0}`

Add 9 both sides:

`\displaystyle{\left(x-4\right)^2-9+9=0+9}\\\\\displaystyle{\left(x-4\right)^2=9}`

Square root both sides:

`\displaystyle{√(\left(x-4\right)^2)=√(9)}`

Cancel square root for left side and add plus-minus to right side:

`\displaystyle{x-4=\pm √(9)}`

Evaluate the surd of 9:

`\displaystyle{x-4=\pm 3}`

Add 4 both sides:

`\displaystyle{x-4+4=\pm 3 +4}\\\\\displaystyle{x=\pm 3 +4}`

Two solutions can be either:

`\displaystyle{x=3+4}` or
`\displaystyle{x = -3+4}`

Hence:

`\displaystyle{x=7,1}`