Solve by Factoring (2x-3)^2=25

${(2 x - 3)}^{2} = 25$

Step 1

Subtract $25$ from both sides of the equation.

${(2 x - 3)}^{2} - 25 = 0$

Step 2

Simplify ${(2 x - 3)}^{2} - 25$ .

Simplify each term.

Rewrite ${(2 x - 3)}^{2}$ as $(2 x - 3) (2 x - 3)$ .

$(2 x - 3) (2 x - 3) - 25 = 0$

Expand $(2 x - 3) (2 x - 3)$ using the FOIL Method.

Apply the distributive property.

$2 x (2 x - 3) - 3 (2 x - 3) - 25 = 0$

Apply the distributive property.

$2 x (2 x) + 2 x \cdot - 3 - 3 (2 x - 3) - 25 = 0$

Apply the distributive property.

$2 x (2 x) + 2 x \cdot - 3 - 3 (2 x) - 3 \cdot - 3 - 25 = 0$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$2 \cdot 2 x \cdot x + 2 x \cdot - 3 - 3 (2 x) - 3 \cdot - 3 - 25 = 0$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$2 \cdot 2 (x \cdot x) + 2 x \cdot - 3 - 3 (2 x) - 3 \cdot - 3 - 25 = 0$

Multiply $x$ by $x$ .

$2 \cdot 2 x^{2} + 2 x \cdot - 3 - 3 (2 x) - 3 \cdot - 3 - 25 = 0$

Multiply $2$ by $2$ .

$4 x^{2} + 2 x \cdot - 3 - 3 (2 x) - 3 \cdot - 3 - 25 = 0$

Multiply $- 3$ by $2$ .

$4 x^{2} - 6 x - 3 (2 x) - 3 \cdot - 3 - 25 = 0$

Multiply $2$ by $- 3$ .

$4 x^{2} - 6 x - 6 x - 3 \cdot - 3 - 25 = 0$

Multiply $- 3$ by $- 3$ .

$4 x^{2} - 6 x - 6 x + 9 - 25 = 0$

Subtract $6 x$ from $- 6 x$ .

$4 x^{2} - 12 x + 9 - 25 = 0$

Subtract $25$ from $9$ .

$4 x^{2} - 12 x - 16 = 0$

Step 3

Factor $4$ out of $4 x^{2} - 12 x - 16$ .

Factor $4$ out of $4 x^{2}$ .

$4 (x^{2}) - 12 x - 16 = 0$

Factor $4$ out of $- 12 x$ .

$4 (x^{2}) + 4 (- 3 x) - 16 = 0$

Factor $4$ out of $- 16$ .

$4 x^{2} + 4 (- 3 x) + 4 \cdot - 4 = 0$

Factor $4$ out of $4 x^{2} + 4 (- 3 x)$ .

$4 (x^{2} - 3 x) + 4 \cdot - 4 = 0$

Factor $4$ out of $4 (x^{2} - 3 x) + 4 \cdot - 4$ .

$4 (x^{2} - 3 x - 4) = 0$

Step 4

Factor.

Factor $x^{2} - 3 x - 4$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 4$ and whose sum is $- 3$ .

$- 4, 1$

Write the factored form using these integers.

$4 ((x - 4) (x + 1)) = 0$

Remove unnecessary parentheses.

$4 (x - 4) (x + 1) = 0$

Step 5

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 4 = 0$

$x + 1 = 0$

Step 6

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Step 7

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 8

The final solution is all the values that make $4 (x - 4) (x + 1) = 0$ true.

$x = 4, - 1$

Solve by Factoring (2x-3)^2=25

Related Questions

Support LCMGCF.com by disabling your adblocker.