Solve 20k^2-98k+72

$$20 k ^ { 2 } - 98 k + 72$$

Factor

$$2\left(k-4\right)\left(10k-9\right)$$

Solution Steps

Factor out 2.

$$2\left(10k^{2}-49k+36\right)$$

Consider 10k^{2}-49k+36. Factor the expression by grouping. First, the expression needs to be rewritten as 10k^{2}+ak+bk+36. To find a and b, set up a system to be solved.

$$a+b=-49 ab=10\times 36=360$$

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 360.

$$-1,-360 -2,-180 -3,-120 -4,-90 -5,-72 -6,-60 -8,-45 -9,-40 -10,-36 -12,-30 -15,-24 -18,-20$$

Calculate the sum for each pair.

$$-1-360=-361 -2-180=-182 -3-120=-123 -4-90=-94 -5-72=-77 -6-60=-66 -8-45=-53 -9-40=-49 -10-36=-46 -12-30=-42 -15-24=-39 -18-20=-38$$

The solution is the pair that gives sum -49.

$$a=-40 b=-9$$

Rewrite 10k^{2}-49k+36 as \left(10k^{2}-40k\right)+\left(-9k+36\right).

$$\left(10k^{2}-40k\right)+\left(-9k+36\right)$$

Factor out 10k in the first and -9 in the second group.

$$10k\left(k-4\right)-9\left(k-4\right)$$

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

$$\left(k-4\right)\left(10k-9\right)$$

Rewrite the complete factored expression.

$$2\left(k-4\right)\left(10k-9\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$20k^{2}-98k+72=0$$

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.

$$k=\frac{-\left(-98\right)±\sqrt{\left(-98\right)^{2}-4\times 20\times 72}}{2\times 20}$$

Square -98.

$$k=\frac{-\left(-98\right)±\sqrt{9604-4\times 20\times 72}}{2\times 20}$$

Multiply -4 times 20.

$$k=\frac{-\left(-98\right)±\sqrt{9604-80\times 72}}{2\times 20}$$

Multiply -80 times 72.

$$k=\frac{-\left(-98\right)±\sqrt{9604-5760}}{2\times 20}$$

Add 9604 to -5760.

$$k=\frac{-\left(-98\right)±\sqrt{3844}}{2\times 20}$$

Take the square root of 3844.

$$k=\frac{-\left(-98\right)±62}{2\times 20}$$

The opposite of -98 is 98.

$$k=\frac{98±62}{2\times 20}$$

Multiply 2 times 20.

$$k=\frac{98±62}{40}$$

Now solve the equation k=\frac{98±62}{40} when ± is plus. Add 98 to 62.

$$k=\frac{160}{40}$$

Divide 160 by 40.

$$k=4$$

Now solve the equation k=\frac{98±62}{40} when ± is minus. Subtract 62 from 98.

$$k=\frac{36}{40}$$

Reduce the fraction \frac{36}{40} to lowest terms by extracting and canceling out 4.

$$k=\frac{9}{10}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 4 for x_{1} and \frac{9}{10} for x_{2}.

$$20k^{2}-98k+72=20\left(k-4\right)\left(k-\frac{9}{10}\right)$$

Subtract \frac{9}{10} from k by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$20k^{2}-98k+72=20\left(k-4\right)\times \frac{10k-9}{10}$$

Cancel out 10, the greatest common factor in 20 and 10.

$$20k^{2}-98k+72=2\left(k-4\right)\left(10k-9\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 20

$$x ^ 2 -\frac{49}{10}x +\frac{18}{5} = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = \frac{49}{10} rs = \frac{18}{5}$$

Two numbers r and s sum up to \frac{49}{10} exactly when the average of the two numbers is \frac{1}{2}*\frac{49}{10} = \frac{49}{20}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = \frac{49}{20} - u s = \frac{49}{20} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{18}{5}

$$(\frac{49}{20} - u) (\frac{49}{20} + u) = \frac{18}{5}$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{2401}{400} - u^2 = \frac{18}{5}$$

Simplify the expression by subtracting \frac{2401}{400} on both sides

$$-u^2 = \frac{18}{5}-\frac{2401}{400} = -\frac{961}{400}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{961}{400} u = \pm\sqrt{\frac{961}{400}} = \pm \frac{31}{20}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =\frac{49}{20} - \frac{31}{20} = 0.900 s = \frac{49}{20} + \frac{31}{20} = 4.000$$

Evaluate

$$2\left(k-4\right)\left(10k-9\right)$$