Solve 24k^2-50k=-24

$$24 k ^ { 2 } - 50 k = - 24$$

Solve for k

$$k=\frac{3}{4}=0.75$$

$$k = \frac{4}{3} = 1\frac{1}{3} \approx 1.333333333$$

Steps Using Factoring By Grouping

Add 24 to both sides.

$$24k^{2}-50k+24=0$$

Divide both sides by 2.

$$12k^{2}-25k+12=0$$

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

$$a+b=-25 ab=12\times 12=144$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 144.

$$-1,-144 -2,-72 -3,-48 -4,-36 -6,-24 -8,-18 -9,-16 -12,-12$$

Calculate the sum for each pair.

$$-1-144=-145 -2-72=-74 -3-48=-51 -4-36=-40 -6-24=-30 -8-18=-26 -9-16=-25 -12-12=-24$$

The solution is the pair that gives sum -25.

$$a=-16 b=-9$$

Rewrite 12k^{2}-25k+12 as \left(12k^{2}-16k\right)+\left(-9k+12\right).

$$\left(12k^{2}-16k\right)+\left(-9k+12\right)$$

Factor out 4k in the first and -3 in the second group.

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

Factor out common term 3k-4 by using distributive property.

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

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

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

Steps Using the Quadratic Formula

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.

$$24k^{2}-50k=-24$$

Add 24 to both sides of the equation.

$$24k^{2}-50k-\left(-24\right)=-24-\left(-24\right)$$

Subtracting -24 from itself leaves 0.

$$24k^{2}-50k-\left(-24\right)=0$$

Subtract -24 from 0.

$$24k^{2}-50k+24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 24 for a, -50 for b, and 24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-\left(-50\right)±\sqrt{\left(-50\right)^{2}-4\times 24\times 24}}{2\times 24}$$

Square -50.

$$k=\frac{-\left(-50\right)±\sqrt{2500-4\times 24\times 24}}{2\times 24}$$

Multiply -4 times 24.

$$k=\frac{-\left(-50\right)±\sqrt{2500-96\times 24}}{2\times 24}$$

Multiply -96 times 24.

$$k=\frac{-\left(-50\right)±\sqrt{2500-2304}}{2\times 24}$$

Add 2500 to -2304.

$$k=\frac{-\left(-50\right)±\sqrt{196}}{2\times 24}$$

Take the square root of 196.

$$k=\frac{-\left(-50\right)±14}{2\times 24}$$

The opposite of -50 is 50.

$$k=\frac{50±14}{2\times 24}$$

Multiply 2 times 24.

$$k=\frac{50±14}{48}$$

Now solve the equation k=\frac{50±14}{48} when ± is plus. Add 50 to 14.

$$k=\frac{64}{48}$$

Reduce the fraction \frac{64}{48} to lowest terms by extracting and canceling out 16.

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

Now solve the equation k=\frac{50±14}{48} when ± is minus. Subtract 14 from 50.

$$k=\frac{36}{48}$$

Reduce the fraction \frac{36}{48} to lowest terms by extracting and canceling out 12.

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

The equation is now solved.

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

Steps for Completing the Square

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.

$$24k^{2}-50k=-24$$

Divide both sides by 24.

$$\frac{24k^{2}-50k}{24}=-\frac{24}{24}$$

Dividing by 24 undoes the multiplication by 24.

$$k^{2}+\left(-\frac{50}{24}\right)k=-\frac{24}{24}$$

Reduce the fraction \frac{-50}{24} to lowest terms by extracting and canceling out 2.

$$k^{2}-\frac{25}{12}k=-\frac{24}{24}$$

Divide -24 by 24.

$$k^{2}-\frac{25}{12}k=-1$$

Divide -\frac{25}{12}, the coefficient of the x term, by 2 to get -\frac{25}{24}. Then add the square of -\frac{25}{24} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$k^{2}-\frac{25}{12}k+\left(-\frac{25}{24}\right)^{2}=-1+\left(-\frac{25}{24}\right)^{2}$$

Square -\frac{25}{24} by squaring both the numerator and the denominator of the fraction.

$$k^{2}-\frac{25}{12}k+\frac{625}{576}=-1+\frac{625}{576}$$

Add -1 to \frac{625}{576}.

$$k^{2}-\frac{25}{12}k+\frac{625}{576}=\frac{49}{576}$$

Factor k^{2}-\frac{25}{12}k+\frac{625}{576}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(k-\frac{25}{24}\right)^{2}=\frac{49}{576}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k-\frac{25}{24}\right)^{2}}=\sqrt{\frac{49}{576}}$$

Simplify.

$$k-\frac{25}{24}=\frac{7}{24} k-\frac{25}{24}=-\frac{7}{24}$$

Add \frac{25}{24} to both sides of the equation.

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