Solve 9x^2-68x+84

$$9 x ^ { 2 } - 68 x + 84$$

Factor

$$\left(x-6\right)\left(9x-14\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 9x^{2}+ax+bx+84. To find a and b, set up a system to be solved.

$$a+b=-68 ab=9\times 84=756$$

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 756.

$$-1,-756 -2,-378 -3,-252 -4,-189 -6,-126 -7,-108 -9,-84 -12,-63 -14,-54 -18,-42 -21,-36 -27,-28$$

Calculate the sum for each pair.

$$-1-756=-757 -2-378=-380 -3-252=-255 -4-189=-193 -6-126=-132 -7-108=-115 -9-84=-93 -12-63=-75 -14-54=-68 -18-42=-60 -21-36=-57 -27-28=-55$$

The solution is the pair that gives sum -68.

$$a=-54 b=-14$$

Rewrite 9x^{2}-68x+84 as \left(9x^{2}-54x\right)+\left(-14x+84\right).

$$\left(9x^{2}-54x\right)+\left(-14x+84\right)$$

Factor out 9x in the first and -14 in the second group.

$$9x\left(x-6\right)-14\left(x-6\right)$$

Factor out common term x-6 by using distributive property.

$$\left(x-6\right)\left(9x-14\right)$$

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.

$$9x^{2}-68x+84=0$$

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.

$$x=\frac{-\left(-68\right)±\sqrt{\left(-68\right)^{2}-4\times 9\times 84}}{2\times 9}$$

Square -68.

$$x=\frac{-\left(-68\right)±\sqrt{4624-4\times 9\times 84}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-\left(-68\right)±\sqrt{4624-36\times 84}}{2\times 9}$$

Multiply -36 times 84.

$$x=\frac{-\left(-68\right)±\sqrt{4624-3024}}{2\times 9}$$

Add 4624 to -3024.

$$x=\frac{-\left(-68\right)±\sqrt{1600}}{2\times 9}$$

Take the square root of 1600.

$$x=\frac{-\left(-68\right)±40}{2\times 9}$$

The opposite of -68 is 68.

$$x=\frac{68±40}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{68±40}{18}$$

Now solve the equation x=\frac{68±40}{18} when ± is plus. Add 68 to 40.

$$x=\frac{108}{18}$$

Divide 108 by 18.

$$x=6$$

Now solve the equation x=\frac{68±40}{18} when ± is minus. Subtract 40 from 68.

$$x=\frac{28}{18}$$

Reduce the fraction \frac{28}{18} to lowest terms by extracting and canceling out 2.

$$x=\frac{14}{9}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 6 for x_{1} and \frac{14}{9} for x_{2}.

$$9x^{2}-68x+84=9\left(x-6\right)\left(x-\frac{14}{9}\right)$$

Subtract \frac{14}{9} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$9x^{2}-68x+84=9\left(x-6\right)\times \frac{9x-14}{9}$$

Cancel out 9, the greatest common factor in 9 and 9.

$$9x^{2}-68x+84=\left(x-6\right)\left(9x-14\right)$$

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 9

$$x ^ 2 -\frac{68}{9}x +\frac{28}{3} = 0$$

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

$$r + s = \frac{68}{9} rs = \frac{28}{3}$$

Two numbers r and s sum up to \frac{68}{9} exactly when the average of the two numbers is \frac{1}{2}*\frac{68}{9} = \frac{34}{9}. 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>

$$r = \frac{34}{9} - u s = \frac{34}{9} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{28}{3}

$$(\frac{34}{9} - u) (\frac{34}{9} + u) = \frac{28}{3}$$

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

$$\frac{1156}{81} - u^2 = \frac{28}{3}$$

Simplify the expression by subtracting \frac{1156}{81} on both sides

$$-u^2 = \frac{28}{3}-\frac{1156}{81} = -\frac{400}{81}$$

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

$$u^2 = \frac{400}{81} u = \pm\sqrt{\frac{400}{81}} = \pm \frac{20}{9}$$

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

$$r =\frac{34}{9} - \frac{20}{9} = 1.556 s = \frac{34}{9} + \frac{20}{9} = 6$$

Evaluate

$$\left(x-6\right)\left(9x-14\right)$$