Solve 3-k/4-k=k-3

$$\frac { 3 - k } { 4 - k } = k - 3$$

Solve for k

$$k=3$$

$$k=5$$

Steps Using the Quadratic Formula

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

$$3-k=\left(-k+4\right)k+\left(-k+4\right)\left(-3\right)$$

Use the distributive property to multiply -k+4 by k.

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

Use the distributive property to multiply -k+4 by -3.

$$3-k=-k^{2}+4k+3k-12$$

Combine 4k and 3k to get 7k.

$$3-k=-k^{2}+7k-12$$

Add k^{2} to both sides.

$$3-k+k^{2}=7k-12$$

Subtract 7k from both sides.

$$3-k+k^{2}-7k=-12$$

Combine -k and -7k to get -8k.

$$3-8k+k^{2}=-12$$

Add 12 to both sides.

$$3-8k+k^{2}+12=0$$

Add 3 and 12 to get 15.

$$15-8k+k^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$k^{2}-8k+15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -8 for b, and 15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 15}}{2}$$

Square -8.

$$k=\frac{-\left(-8\right)±\sqrt{64-4\times 15}}{2}$$

Multiply -4 times 15.

$$k=\frac{-\left(-8\right)±\sqrt{64-60}}{2}$$

Add 64 to -60.

$$k=\frac{-\left(-8\right)±\sqrt{4}}{2}$$

Take the square root of 4.

$$k=\frac{-\left(-8\right)±2}{2}$$

The opposite of -8 is 8.

$$k=\frac{8±2}{2}$$

Now solve the equation k=\frac{8±2}{2} when ± is plus. Add 8 to 2.

$$k=\frac{10}{2}$$

Divide 10 by 2.

$$k=5$$

Now solve the equation k=\frac{8±2}{2} when ± is minus. Subtract 2 from 8.

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

Divide 6 by 2.

$$k=3$$

The equation is now solved.

$$k=5 k=3$$

Steps for Completing the Square

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

$$3-k=\left(-k+4\right)k+\left(-k+4\right)\left(-3\right)$$

Use the distributive property to multiply -k+4 by k.

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

Use the distributive property to multiply -k+4 by -3.

$$3-k=-k^{2}+4k+3k-12$$

Combine 4k and 3k to get 7k.

$$3-k=-k^{2}+7k-12$$

Add k^{2} to both sides.

$$3-k+k^{2}=7k-12$$

Subtract 7k from both sides.

$$3-k+k^{2}-7k=-12$$

Combine -k and -7k to get -8k.

$$3-8k+k^{2}=-12$$

Subtract 3 from both sides.

$$-8k+k^{2}=-12-3$$

Subtract 3 from -12 to get -15.

$$-8k+k^{2}=-15$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$k^{2}-8k=-15$$

Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$k^{2}-8k+\left(-4\right)^{2}=-15+\left(-4\right)^{2}$$

Square -4.

$$k^{2}-8k+16=-15+16$$

Add -15 to 16.

$$k^{2}-8k+16=1$$

Factor k^{2}-8k+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(k-4\right)^{2}=1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k-4\right)^{2}}=\sqrt{1}$$

Simplify.

$$k-4=1 k-4=-1$$

Add 4 to both sides of the equation.

$$k=5 k=3$$