Solve q^2-69q+68

$$q ^ { 2 } - 69 q + 68$$

Factor

$$\left(q-68\right)\left(q-1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as q^{2}+aq+bq+68. To find a and b, set up a system to be solved.

$$a+b=-69 ab=1\times 68=68$$

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 68.

$$-1,-68 -2,-34 -4,-17$$

Calculate the sum for each pair.

$$-1-68=-69 -2-34=-36 -4-17=-21$$

The solution is the pair that gives sum -69.

$$a=-68 b=-1$$

Rewrite q^{2}-69q+68 as \left(q^{2}-68q\right)+\left(-q+68\right).

$$\left(q^{2}-68q\right)+\left(-q+68\right)$$

Factor out q in the first and -1 in the second group.

$$q\left(q-68\right)-\left(q-68\right)$$

Factor out common term q-68 by using distributive property.

$$\left(q-68\right)\left(q-1\right)$$

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.

$$q^{2}-69q+68=0$$

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.

$$q=\frac{-\left(-69\right)±\sqrt{\left(-69\right)^{2}-4\times 68}}{2}$$

Square -69.

$$q=\frac{-\left(-69\right)±\sqrt{4761-4\times 68}}{2}$$

Multiply -4 times 68.

$$q=\frac{-\left(-69\right)±\sqrt{4761-272}}{2}$$

Add 4761 to -272.

$$q=\frac{-\left(-69\right)±\sqrt{4489}}{2}$$

Take the square root of 4489.

$$q=\frac{-\left(-69\right)±67}{2}$$

The opposite of -69 is 69.

$$q=\frac{69±67}{2}$$

Now solve the equation q=\frac{69±67}{2} when ± is plus. Add 69 to 67.

$$q=\frac{136}{2}$$

Divide 136 by 2.

$$q=68$$

Now solve the equation q=\frac{69±67}{2} when ± is minus. Subtract 67 from 69.

$$q=\frac{2}{2}$$

Divide 2 by 2.

$$q=1$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 68 for x_{1} and 1 for x_{2}.

$$q^{2}-69q+68=\left(q-68\right)\left(q-1\right)$$

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.

$$x ^ 2 -69x +68 = 0$$

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

$$r + s = 69 rs = 68$$

Two numbers r and s sum up to 69 exactly when the average of the two numbers is \frac{1}{2}*69 = \frac{69}{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>

$$r = \frac{69}{2} - u s = \frac{69}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 68

$$(\frac{69}{2} - u) (\frac{69}{2} + u) = 68$$

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

$$\frac{4761}{4} - u^2 = 68$$

Simplify the expression by subtracting \frac{4761}{4} on both sides

$$-u^2 = 68-\frac{4761}{4} = -\frac{4489}{4}$$

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

$$u^2 = \frac{4489}{4} u = \pm\sqrt{\frac{4489}{4}} = \pm \frac{67}{2}$$

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

$$r =\frac{69}{2} - \frac{67}{2} = 1 s = \frac{69}{2} + \frac{67}{2} = 68$$

Evaluate

$$\left(q-68\right)\left(q-1\right)$$