Notice that this time the left-hand side contains a sum of squares. However, we can use the difference of two squares by subtracting one of the square terms from both sides of the equation. Let's do it!

(x + 2)^{2} + (x - 2)^{2} = (2 x)^{2}

SubEqn

$LHS - (x - 2)^{2} = RHS - (x - 2)^{2}$

(x + 2)^{2} = (2 x)^{2} - (x - 2)^{2}

FacDiffSquares

$a^{2} - b^{2} = (a + b) (a - b)$

(x + 2)^{2} = (2 x + (x - 2)) (2 x - (x - 2))

▼

Simplify right-hand side

Distr

Distribute $- 1$

(x + 2)^{2} = (2 x + (x - 2)) (2 x - x + 2)

AddSubTerms

Add and subtract terms

(x + 2)^{2} = (3 x - 2) (x + 2)

Now, notice that both sides of the equation have the

(x + 2)

term. Therefore, we can subtract

(3 x - 2) (x + 2)

from both sides of the equation and use the Zero Product Property to change our quadratic equation into two linear ones.

(x + 2)^{2} = (3 x - 2) (x + 2)

SubEqn

$LHS - (3 x + 2) (x + 2) = RHS - (3 x + 2) (x + 2)$

(x + 2)^{2} - (3 x - 2) (x + 2) = 0

FactorOut

Factor out $(x + 2)$

(x + 2) ((x + 2) - (3 x - 2)) = 0

▼

Simplify left-hand side

Distr

Distribute $- 1$

(x + 2) (x + 2 - 3 x + 2) = 0

SubTerms

Subtract terms

(x + 2) (- 2 x + 4) = 0

FactorOut

Factor out $(- 2)$

- 2 (x + 2) (x - 2) = 0

DivEqn

$LHS / (- 2) = RHS / (- 2)$

(x + 2) (x - 2) = 0

SubEqn

$(I):$ $LHS - 2 = RHS - 2$

x = - x - 2 = 0 (I) (II)

AddEqn

$(II):$ $LHS + 2 = RHS + 2$

x = - 2 x = 2

Therefore, our equation has two solutions:

x = - 2

and

x = 2 .

Subchapter links

11.2.1 Creating a Three-Dimensional Model p.582-583

11.2.2 Graphing Equations in Three Dimensions p.586-587

11.2.3 Solving Systems of Three Equations with Three Variables p.590-592

11.2.4 Using Systems of Three Equations for Curve Fitting p.598-600

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