Solve 51x^2-64x+20

$$51 x ^ { 2 } - 64 x + 20$$

Factor

$$\left(3x-2\right)\left(17x-10\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 51x^{2}+ax+bx+20. To find a and b, set up a system to be solved.

$$a+b=-64 ab=51\times 20=1020$$

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 1020.

$$-1,-1020 -2,-510 -3,-340 -4,-255 -5,-204 -6,-170 -10,-102 -12,-85 -15,-68 -17,-60 -20,-51 -30,-34$$

Calculate the sum for each pair.

$$-1-1020=-1021 -2-510=-512 -3-340=-343 -4-255=-259 -5-204=-209 -6-170=-176 -10-102=-112 -12-85=-97 -15-68=-83 -17-60=-77 -20-51=-71 -30-34=-64$$

The solution is the pair that gives sum -64.

$$a=-34 b=-30$$

Rewrite 51x^{2}-64x+20 as \left(51x^{2}-34x\right)+\left(-30x+20\right).

$$\left(51x^{2}-34x\right)+\left(-30x+20\right)$$

Factor out 17x in the first and -10 in the second group.

$$17x\left(3x-2\right)-10\left(3x-2\right)$$

Factor out common term 3x-2 by using distributive property.

$$\left(3x-2\right)\left(17x-10\right)$$

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.

$$51x^{2}-64x+20=0$$

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.

$$x=\frac{-\left(-64\right)±\sqrt{\left(-64\right)^{2}-4\times 51\times 20}}{2\times 51}$$

Square -64.

$$x=\frac{-\left(-64\right)±\sqrt{4096-4\times 51\times 20}}{2\times 51}$$

Multiply -4 times 51.

$$x=\frac{-\left(-64\right)±\sqrt{4096-204\times 20}}{2\times 51}$$

Multiply -204 times 20.

$$x=\frac{-\left(-64\right)±\sqrt{4096-4080}}{2\times 51}$$

Add 4096 to -4080.

$$x=\frac{-\left(-64\right)±\sqrt{16}}{2\times 51}$$

Take the square root of 16.

$$x=\frac{-\left(-64\right)±4}{2\times 51}$$

The opposite of -64 is 64.

$$x=\frac{64±4}{2\times 51}$$

Multiply 2 times 51.

$$x=\frac{64±4}{102}$$

Now solve the equation x=\frac{64±4}{102} when ± is plus. Add 64 to 4.

$$x=\frac{68}{102}$$

Reduce the fraction \frac{68}{102} to lowest terms by extracting and canceling out 34.

$$x=\frac{2}{3}$$

Now solve the equation x=\frac{64±4}{102} when ± is minus. Subtract 4 from 64.

$$x=\frac{60}{102}$$

Reduce the fraction \frac{60}{102} to lowest terms by extracting and canceling out 6.

$$x=\frac{10}{17}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{2}{3} for x_{1} and \frac{10}{17} for x_{2}.

$$51x^{2}-64x+20=51\left(x-\frac{2}{3}\right)\left(x-\frac{10}{17}\right)$$

Subtract \frac{2}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$51x^{2}-64x+20=51\times \frac{3x-2}{3}\left(x-\frac{10}{17}\right)$$

Subtract \frac{10}{17} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$51x^{2}-64x+20=51\times \frac{3x-2}{3}\times \frac{17x-10}{17}$$

Multiply \frac{3x-2}{3} times \frac{17x-10}{17} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$51x^{2}-64x+20=51\times \frac{\left(3x-2\right)\left(17x-10\right)}{3\times 17}$$

Multiply 3 times 17.

$$51x^{2}-64x+20=51\times \frac{\left(3x-2\right)\left(17x-10\right)}{51}$$

Cancel out 51, the greatest common factor in 51 and 51.

$$51x^{2}-64x+20=\left(3x-2\right)\left(17x-10\right)$$

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 51

$$x ^ 2 -\frac{64}{51}x +\frac{20}{51} = 0$$

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

$$r + s = \frac{64}{51} rs = \frac{20}{51}$$

Two numbers r and s sum up to \frac{64}{51} exactly when the average of the two numbers is \frac{1}{2}*\frac{64}{51} = \frac{32}{51}. 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-gzdabgg4ehffg0hf.b01.azurefd.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = \frac{32}{51} - u s = \frac{32}{51} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{20}{51}

$$(\frac{32}{51} - u) (\frac{32}{51} + u) = \frac{20}{51}$$

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

$$\frac{1024}{2601} - u^2 = \frac{20}{51}$$

Simplify the expression by subtracting \frac{1024}{2601} on both sides

$$-u^2 = \frac{20}{51}-\frac{1024}{2601} = -\frac{4}{2601}$$

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

$$u^2 = \frac{4}{2601} u = \pm\sqrt{\frac{4}{2601}} = \pm \frac{2}{51}$$

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

$$r =\frac{32}{51} - \frac{2}{51} = 0.588 s = \frac{32}{51} + \frac{2}{51} = 0.667$$

Evaluate

$$\left(3x-2\right)\left(17x-10\right)$$