Solve -14x^2+133x-63

$$- 14 x ^ { 2 } + 133 x - 63$$

Factor

$$7\left(9-x\right)\left(2x-1\right)$$

Solution Steps

Factor out 7.

$$7\left(-2x^{2}+19x-9\right)$$

Consider -2x^{2}+19x-9. Factor the expression by grouping. First, the expression needs to be rewritten as -2x^{2}+ax+bx-9. To find a and b, set up a system to be solved.

$$a+b=19 ab=-2\left(-9\right)=18$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 18.

$$1,18 2,9 3,6$$

Calculate the sum for each pair.

$$1+18=19 2+9=11 3+6=9$$

The solution is the pair that gives sum 19.

$$a=18 b=1$$

Rewrite -2x^{2}+19x-9 as \left(-2x^{2}+18x\right)+\left(x-9\right).

$$\left(-2x^{2}+18x\right)+\left(x-9\right)$$

Factor out 2x in the first and -1 in the second group.

$$2x\left(-x+9\right)-\left(-x+9\right)$$

Factor out common term -x+9 by using distributive property.

$$\left(-x+9\right)\left(2x-1\right)$$

Rewrite the complete factored expression.

$$7\left(-x+9\right)\left(2x-1\right)$$

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.

$$-14x^{2}+133x-63=0$$

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.

$$x=\frac{-133±\sqrt{133^{2}-4\left(-14\right)\left(-63\right)}}{2\left(-14\right)}$$

Square 133.

$$x=\frac{-133±\sqrt{17689-4\left(-14\right)\left(-63\right)}}{2\left(-14\right)}$$

Multiply -4 times -14.

$$x=\frac{-133±\sqrt{17689+56\left(-63\right)}}{2\left(-14\right)}$$

Multiply 56 times -63.

$$x=\frac{-133±\sqrt{17689-3528}}{2\left(-14\right)}$$

Add 17689 to -3528.

$$x=\frac{-133±\sqrt{14161}}{2\left(-14\right)}$$

Take the square root of 14161.

$$x=\frac{-133±119}{2\left(-14\right)}$$

Multiply 2 times -14.

$$x=\frac{-133±119}{-28}$$

Now solve the equation x=\frac{-133±119}{-28} when ± is plus. Add -133 to 119.

$$x=-\frac{14}{-28}$$

Reduce the fraction \frac{-14}{-28} to lowest terms by extracting and canceling out 14.

$$x=\frac{1}{2}$$

Now solve the equation x=\frac{-133±119}{-28} when ± is minus. Subtract 119 from -133.

$$x=-\frac{252}{-28}$$

Divide -252 by -28.

$$x=9$$

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

$$-14x^{2}+133x-63=-14\left(x-\frac{1}{2}\right)\left(x-9\right)$$

Subtract \frac{1}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-14x^{2}+133x-63=-14\times \frac{-2x+1}{-2}\left(x-9\right)$$

Cancel out 2, the greatest common factor in -14 and 2.

$$-14x^{2}+133x-63=7\left(-2x+1\right)\left(x-9\right)$$

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.

$$x ^ 2 -\frac{19}{2}x +\frac{9}{2} = 0$$

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

$$r + s = \frac{19}{2} rs = \frac{9}{2}$$

Two numbers r and s sum up to \frac{19}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{19}{2} = \frac{19}{4}. 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>

$$r = \frac{19}{4} - u s = \frac{19}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{9}{2}

$$(\frac{19}{4} - u) (\frac{19}{4} + u) = \frac{9}{2}$$

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

$$\frac{361}{16} - u^2 = \frac{9}{2}$$

Simplify the expression by subtracting \frac{361}{16} on both sides

$$-u^2 = \frac{9}{2}-\frac{361}{16} = -\frac{289}{16}$$

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

$$u^2 = \frac{289}{16} u = \pm\sqrt{\frac{289}{16}} = \pm \frac{17}{4}$$

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

$$r =\frac{19}{4} - \frac{17}{4} = 0.500 s = \frac{19}{4} + \frac{17}{4} = 9$$

Evaluate

$$-14x^{2}+133x-63$$