Solve n^2-19n+78

$$n ^ { 2 } - 19 n + 78$$

Factor

$$\left(n-13\right)\left(n-6\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as n^{2}+an+bn+78. To find a and b, set up a system to be solved.

$$a+b=-19 ab=1\times 78=78$$

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 78.

$$-1,-78 -2,-39 -3,-26 -6,-13$$

Calculate the sum for each pair.

$$-1-78=-79 -2-39=-41 -3-26=-29 -6-13=-19$$

The solution is the pair that gives sum -19.

$$a=-13 b=-6$$

Rewrite n^{2}-19n+78 as \left(n^{2}-13n\right)+\left(-6n+78\right).

$$\left(n^{2}-13n\right)+\left(-6n+78\right)$$

Factor out n in the first and -6 in the second group.

$$n\left(n-13\right)-6\left(n-13\right)$$

Factor out common term n-13 by using distributive property.

$$\left(n-13\right)\left(n-6\right)$$

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.

$$n^{2}-19n+78=0$$

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.

$$n=\frac{-\left(-19\right)±\sqrt{\left(-19\right)^{2}-4\times 78}}{2}$$

Square -19.

$$n=\frac{-\left(-19\right)±\sqrt{361-4\times 78}}{2}$$

Multiply -4 times 78.

$$n=\frac{-\left(-19\right)±\sqrt{361-312}}{2}$$

Add 361 to -312.

$$n=\frac{-\left(-19\right)±\sqrt{49}}{2}$$

Take the square root of 49.

$$n=\frac{-\left(-19\right)±7}{2}$$

The opposite of -19 is 19.

$$n=\frac{19±7}{2}$$

Now solve the equation n=\frac{19±7}{2} when ± is plus. Add 19 to 7.

$$n=\frac{26}{2}$$

Divide 26 by 2.

$$n=13$$

Now solve the equation n=\frac{19±7}{2} when ± is minus. Subtract 7 from 19.

$$n=\frac{12}{2}$$

Divide 12 by 2.

$$n=6$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 13 for x_{1} and 6 for x_{2}.

$$n^{2}-19n+78=\left(n-13\right)\left(n-6\right)$$

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.

$$x ^ 2 -19x +78 = 0$$

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

$$r + s = 19 rs = 78$$

Two numbers r and s sum up to 19 exactly when the average of the two numbers is \frac{1}{2}*19 = \frac{19}{2}. 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>

$$r = \frac{19}{2} - u s = \frac{19}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 78

$$(\frac{19}{2} - u) (\frac{19}{2} + u) = 78$$

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

$$\frac{361}{4} - u^2 = 78$$

Simplify the expression by subtracting \frac{361}{4} on both sides

$$-u^2 = 78-\frac{361}{4} = -\frac{49}{4}$$

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

$$u^2 = \frac{49}{4} u = \pm\sqrt{\frac{49}{4}} = \pm \frac{7}{2}$$

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

$$r =\frac{19}{2} - \frac{7}{2} = 6 s = \frac{19}{2} + \frac{7}{2} = 13$$

Evaluate

$$\left(n-13\right)\left(n-6\right)$$