Solve x^2-3x+4/x-4=-4

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

Solve for x

$$x=-4$$

$$x=3$$

Steps Using Factoring

Variable x cannot be equal to 4 since division by zero is not defined. Multiply both sides of the equation by x-4.

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

Use the distributive property to multiply -4 by x-4.

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

Add 4x to both sides.

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

Combine -3x and 4x to get x.

$$x^{2}+x+4=16$$

Subtract 16 from both sides.

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

Subtract 16 from 4 to get -12.

$$x^{2}+x-12=0$$

To solve the equation, factor x^{2}+x-12 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=1 ab=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -12.

$$-1,12 -2,6 -3,4$$

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 1.

$$a=-3 b=4$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-3\right)\left(x+4\right)$$

To find equation solutions, solve x-3=0 and x+4=0.

$$x=3 x=-4$$

Steps Using Factoring By Grouping

Variable x cannot be equal to 4 since division by zero is not defined. Multiply both sides of the equation by x-4.

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

Use the distributive property to multiply -4 by x-4.

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

Add 4x to both sides.

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

Combine -3x and 4x to get x.

$$x^{2}+x+4=16$$

Subtract 16 from both sides.

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

Subtract 16 from 4 to get -12.

$$x^{2}+x-12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-12. To find a and b, set up a system to be solved.

$$a+b=1 ab=1\left(-12\right)=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -12.

$$-1,12 -2,6 -3,4$$

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 1.

$$a=-3 b=4$$

Rewrite x^{2}+x-12 as \left(x^{2}-3x\right)+\left(4x-12\right).

$$\left(x^{2}-3x\right)+\left(4x-12\right)$$

Factor out x in the first and 4 in the second group.

$$x\left(x-3\right)+4\left(x-3\right)$$

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(x+4\right)$$

To find equation solutions, solve x-3=0 and x+4=0.

$$x=3 x=-4$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 4 since division by zero is not defined. Multiply both sides of the equation by x-4.

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

Use the distributive property to multiply -4 by x-4.

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

Add 4x to both sides.

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

Combine -3x and 4x to get x.

$$x^{2}+x+4=16$$

Subtract 16 from both sides.

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

Subtract 16 from 4 to get -12.

$$x^{2}+x-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 1 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\left(-12\right)}}{2}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\left(-12\right)}}{2}$$

Multiply -4 times -12.

$$x=\frac{-1±\sqrt{1+48}}{2}$$

Add 1 to 48.

$$x=\frac{-1±\sqrt{49}}{2}$$

Take the square root of 49.

$$x=\frac{-1±7}{2}$$

Now solve the equation x=\frac{-1±7}{2} when ± is plus. Add -1 to 7.

$$x=\frac{6}{2}$$

Divide 6 by 2.

$$x=3$$

Now solve the equation x=\frac{-1±7}{2} when ± is minus. Subtract 7 from -1.

$$x=-\frac{8}{2}$$

Divide -8 by 2.

$$x=-4$$

The equation is now solved.

$$x=3 x=-4$$

Steps for Completing the Square

Variable x cannot be equal to 4 since division by zero is not defined. Multiply both sides of the equation by x-4.

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

Use the distributive property to multiply -4 by x-4.

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

Add 4x to both sides.

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

Combine -3x and 4x to get x.

$$x^{2}+x+4=16$$

Subtract 4 from both sides.

$$x^{2}+x=16-4$$

Subtract 4 from 16 to get 12.

$$x^{2}+x=12$$

Divide 1, the coefficient of the x term, by 2 to get \frac{1}{2}. Then add the square of \frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=12+\left(\frac{1}{2}\right)^{2}$$

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+x+\frac{1}{4}=12+\frac{1}{4}$$

Add 12 to \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=\frac{49}{4}$$

Factor x^{2}+x+\frac{1}{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}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$x+\frac{1}{2}=\frac{7}{2} x+\frac{1}{2}=-\frac{7}{2}$$

Subtract \frac{1}{2} from both sides of the equation.

$$x=3 x=-4$$