Solve (x-1):(x+1)=(2x-4):(x+4)

$$( x - 1 ) : ( x + 1 ) = ( 2 x - 4 ) : ( x + 4 )$$

Solve for x

$$x=5$$

$$x=0$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -4,-1 since division by zero is not defined. Multiply both sides of the equation by \left(x+1\right)\left(x+4\right), the least common multiple of x+1,x+4.

$$\left(x+4\right)\left(x-1\right)=\left(x+1\right)\left(2x-4\right)$$

Use the distributive property to multiply x+4 by x-1 and combine like terms.

$$x^{2}+3x-4=\left(x+1\right)\left(2x-4\right)$$

Use the distributive property to multiply x+1 by 2x-4 and combine like terms.

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

Subtract 2x^{2} from both sides.

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

Combine x^{2} and -2x^{2} to get -x^{2}.

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

Add 2x to both sides.

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

Combine 3x and 2x to get 5x.

$$-x^{2}+5x-4=-4$$

Add 4 to both sides.

$$-x^{2}+5x-4+4=0$$

Add -4 and 4 to get 0.

$$-x^{2}+5x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 5 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}}}{2\left(-1\right)}$$

Take the square root of 5^{2}.

$$x=\frac{-5±5}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-5±5}{-2}$$

Now solve the equation x=\frac{-5±5}{-2} when ± is plus. Add -5 to 5.

$$x=\frac{0}{-2}$$

Divide 0 by -2.

$$x=0$$

Now solve the equation x=\frac{-5±5}{-2} when ± is minus. Subtract 5 from -5.

$$x=-\frac{10}{-2}$$

Divide -10 by -2.

$$x=5$$

The equation is now solved.

$$x=0 x=5$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -4,-1 since division by zero is not defined. Multiply both sides of the equation by \left(x+1\right)\left(x+4\right), the least common multiple of x+1,x+4.

$$\left(x+4\right)\left(x-1\right)=\left(x+1\right)\left(2x-4\right)$$

Use the distributive property to multiply x+4 by x-1 and combine like terms.

$$x^{2}+3x-4=\left(x+1\right)\left(2x-4\right)$$

Use the distributive property to multiply x+1 by 2x-4 and combine like terms.

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

Subtract 2x^{2} from both sides.

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

Combine x^{2} and -2x^{2} to get -x^{2}.

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

Add 2x to both sides.

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

Combine 3x and 2x to get 5x.

$$-x^{2}+5x-4=-4$$

Add 4 to both sides.

$$-x^{2}+5x=-4+4$$

Add -4 and 4 to get 0.

$$-x^{2}+5x=0$$

Divide both sides by -1.

$$\frac{-x^{2}+5x}{-1}=\frac{0}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{5}{-1}x=\frac{0}{-1}$$

Divide 5 by -1.

$$x^{2}-5x=\frac{0}{-1}$$

Divide 0 by -1.

$$x^{2}-5x=0$$

Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-5x+\frac{25}{4}=\frac{25}{4}$$

Factor x^{2}-5x+\frac{25}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$x-\frac{5}{2}=\frac{5}{2} x-\frac{5}{2}=-\frac{5}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$x=5 x=0$$