Question

How to solve this equation?

5(x+1)² + 7(x+3)² = 12(x+2)²

Answer 1

`x=-5`

Explanation:

Solve:

`5(x+1)^2+7(x+3)^2=12(x+2)^2`

Expand using the sum of squares: `(a+b)^2=a^2+2ab+b^2`, or using the FOIL method: https://www.ipracticemath.com/learn/algebra/foil-method-of-binomial-multiplication

`5(x^2+2x+1)+7(x^2+6x+9)=12(x^2+4x+4)`

Expand using the distributive property: `a(b+c)=ab+ac`

`5x^2+10x+5+7x^2+42x+63=12x^2+48x+48`

Collect like terms.

`(5x^2+7x^2)+(10x+42x)+(5+63)=12x^2+48x+48`

Combine like terms.

`12x^2+52x+68=12x^2+48x+48`

Cancel `12x^2` on both sides.

`color(red)cancel(color(black)(12x^2))+52x+68=color(red)cancel(color(black)(12x^2))+48x+48`

Simplify.

`52x+68=48x+48`

Subtract `48x` from both sides.

`52x-48x+68=48`

Simplify.

`4x+68=48`

Subtract `68` from both sides.

`4x=48-68`

Simplify.

`4x=-20`

Divide both sides by `4`.

`x=-20/4`

Simplify.

`x=-5`

Answer 2

`x = -5`

Explanation:

Step by step, performing legal algebraic operations that just happen to simplify the expression:

`5(x+1)^2 + 7(x+3)^2 = 12(x+2)^2`

`5(x^2+2x+1) + 7(x^2+6x+9) = 12(x^2+4x+4)`

`(5x^2+10x+5) + (7x^2+42x+63) = (12x^2+48x+48)`

`(12x^2+52x+68) = (12x^2+48x+48)`

`(12x^2+52x+68) - (12x^2+48x+48) = 0`

`(0x^2+4x+20) = 0`

`4x+20 = 0`

`4x = -20`

`1/cancel(4) * cancel(4)x = 1/cancel(4) * (-cancel(20)^5)`

`x = -5`

I hope this helps,
Steve