Solve for x 3x=x^2-4

$3 x = x^{2} - 4$

Subtract $x^{2}$ from both sides of the equation.

$3 x - x^{2} = - 4$

Add $4$ to both sides of the equation.

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

Factor the left side of the equation.

Factor $- 1$ out of $3 x - x^{2} + 4$ .

Reorder $3 x$ and $- x^{2}$ .

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

Factor $- 1$ out of $- x^{2}$ .

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

Factor $- 1$ out of $3 x$ .

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

Rewrite $4$ as $- 1 (- 4)$ .

$- (x^{2}) - (- 3 x) - 1 \cdot - 4 = 0$

Factor $- 1$ out of $- (x^{2}) - (- 3 x)$ .

$- (x^{2} - 3 x) - 1 \cdot - 4 = 0$

Factor $- 1$ out of $- (x^{2} - 3 x) - 1 (- 4)$ .

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

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.

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

Remove unnecessary parentheses.

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

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

$x - 4 = 0$

$x + 1 = 0$

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.

$x = 4$

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.

$x = - 1$

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

$x = 4, - 1$

Solve for x 3x=x^2-4

Related Questions

Support LCMGCF.com by disabling your adblocker.