# How to solve this equation?

## 5(x+1)² + 7(x+3)² = 12(x+2)²

Jun 23, 2018

$x = - 5$

#### Explanation:

Solve:

$5 {\left(x + 1\right)}^{2} + 7 {\left(x + 3\right)}^{2} = 12 {\left(x + 2\right)}^{2}$

Expand using the sum of squares: ${\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$, or using the FOIL method: https://www.ipracticemath.com/learn/algebra/foil-method-of-binomial-multiplication

$5 \left({x}^{2} + 2 x + 1\right) + 7 \left({x}^{2} + 6 x + 9\right) = 12 \left({x}^{2} + 4 x + 4\right)$

Expand using the distributive property: $a \left(b + c\right) = a b + a c$

$5 {x}^{2} + 10 x + 5 + 7 {x}^{2} + 42 x + 63 = 12 {x}^{2} + 48 x + 48$

Collect like terms.

$\left(5 {x}^{2} + 7 {x}^{2}\right) + \left(10 x + 42 x\right) + \left(5 + 63\right) = 12 {x}^{2} + 48 x + 48$

Combine like terms.

$12 {x}^{2} + 52 x + 68 = 12 {x}^{2} + 48 x + 48$

Cancel $12 {x}^{2}$ on both sides.

$\textcolor{red}{\cancel{\textcolor{b l a c k}{12 {x}^{2}}}} + 52 x + 68 = \textcolor{red}{\cancel{\textcolor{b l a c k}{12 {x}^{2}}}} + 48 x + 48$

Simplify.

$52 x + 68 = 48 x + 48$

Subtract $48 x$ from both sides.

$52 x - 48 x + 68 = 48$

Simplify.

$4 x + 68 = 48$

Subtract $68$ from both sides.

$4 x = 48 - 68$

Simplify.

$4 x = - 20$

Divide both sides by $4$.

$x = - \frac{20}{4}$

Simplify.

$x = - 5$

Jun 23, 2018

$x = - 5$

#### Explanation:

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

$5 {\left(x + 1\right)}^{2} + 7 {\left(x + 3\right)}^{2} = 12 {\left(x + 2\right)}^{2}$

$5 \left({x}^{2} + 2 x + 1\right) + 7 \left({x}^{2} + 6 x + 9\right) = 12 \left({x}^{2} + 4 x + 4\right)$

$\left(5 {x}^{2} + 10 x + 5\right) + \left(7 {x}^{2} + 42 x + 63\right) = \left(12 {x}^{2} + 48 x + 48\right)$

$\left(12 {x}^{2} + 52 x + 68\right) = \left(12 {x}^{2} + 48 x + 48\right)$

$\left(12 {x}^{2} + 52 x + 68\right) - \left(12 {x}^{2} + 48 x + 48\right) = 0$

$\left(0 {x}^{2} + 4 x + 20\right) = 0$

$4 x + 20 = 0$

$4 x = - 20$

$\frac{1}{\cancel{4}} \cdot \cancel{4} x = \frac{1}{\cancel{4}} \cdot \left(- {\cancel{20}}^{5}\right)$

$x = - 5$

I hope this helps,
Steve