Solve 75x^2-100x+25

$$75 x ^ { 2 } - 100 x + 25$$

Factor

$$25\left(x-1\right)\left(3x-1\right)$$

Solution Steps

Factor out 25.

$$25\left(3x^{2}-4x+1\right)$$

Consider 3x^{2}-4x+1. Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=-4 ab=3\times 1=3$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-3 b=-1$$

Rewrite 3x^{2}-4x+1 as \left(3x^{2}-3x\right)+\left(-x+1\right).

$$\left(3x^{2}-3x\right)+\left(-x+1\right)$$

Factor out 3x in the first and -1 in the second group.

$$3x\left(x-1\right)-\left(x-1\right)$$

Factor out common term x-1 by using distributive property.

$$\left(x-1\right)\left(3x-1\right)$$

Rewrite the complete factored expression.

$$25\left(x-1\right)\left(3x-1\right)$$

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.

$$75x^{2}-100x+25=0$$

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.

$$x=\frac{-\left(-100\right)±\sqrt{\left(-100\right)^{2}-4\times 75\times 25}}{2\times 75}$$

Square -100.

$$x=\frac{-\left(-100\right)±\sqrt{10000-4\times 75\times 25}}{2\times 75}$$

Multiply -4 times 75.

$$x=\frac{-\left(-100\right)±\sqrt{10000-300\times 25}}{2\times 75}$$

Multiply -300 times 25.

$$x=\frac{-\left(-100\right)±\sqrt{10000-7500}}{2\times 75}$$

Add 10000 to -7500.

$$x=\frac{-\left(-100\right)±\sqrt{2500}}{2\times 75}$$

Take the square root of 2500.

$$x=\frac{-\left(-100\right)±50}{2\times 75}$$

The opposite of -100 is 100.

$$x=\frac{100±50}{2\times 75}$$

Multiply 2 times 75.

$$x=\frac{100±50}{150}$$

Now solve the equation x=\frac{100±50}{150} when ± is plus. Add 100 to 50.

$$x=\frac{150}{150}$$

Divide 150 by 150.

$$x=1$$

Now solve the equation x=\frac{100±50}{150} when ± is minus. Subtract 50 from 100.

$$x=\frac{50}{150}$$

Reduce the fraction \frac{50}{150} to lowest terms by extracting and canceling out 50.

$$x=\frac{1}{3}$$

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

$$75x^{2}-100x+25=75\left(x-1\right)\left(x-\frac{1}{3}\right)$$

Subtract \frac{1}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$75x^{2}-100x+25=75\left(x-1\right)\times \frac{3x-1}{3}$$

Cancel out 3, the greatest common factor in 75 and 3.

$$75x^{2}-100x+25=25\left(x-1\right)\left(3x-1\right)$$

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.This is achieved by dividing both sides of the equation by 75

$$x ^ 2 -\frac{4}{3}x +\frac{1}{3} = 0$$

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

$$r + s = \frac{4}{3} rs = \frac{1}{3}$$

Two numbers r and s sum up to \frac{4}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{4}{3} = \frac{2}{3}. 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>

$$r = \frac{2}{3} - u s = \frac{2}{3} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{1}{3}

$$(\frac{2}{3} - u) (\frac{2}{3} + u) = \frac{1}{3}$$

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

$$\frac{4}{9} - u^2 = \frac{1}{3}$$

Simplify the expression by subtracting \frac{4}{9} on both sides

$$-u^2 = \frac{1}{3}-\frac{4}{9} = -\frac{1}{9}$$

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

$$u^2 = \frac{1}{9} u = \pm\sqrt{\frac{1}{9}} = \pm \frac{1}{3}$$

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

$$r =\frac{2}{3} - \frac{1}{3} = 0.333 s = \frac{2}{3} + \frac{1}{3} = 1$$

Evaluate

$$25\left(x-1\right)\left(3x-1\right)$$