Solve 20y^2+y-1

$$20 y ^ { 2 } + y - 1$$

Factor

$$\left(5y-1\right)\left(4y+1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 20y^{2}+ay+by-1. To find a and b, set up a system to be solved.

$$a+b=1 ab=20\left(-1\right)=-20$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -20.

$$-1,20 -2,10 -4,5$$

Calculate the sum for each pair.

$$-1+20=19 -2+10=8 -4+5=1$$

The solution is the pair that gives sum 1.

$$a=-4 b=5$$

Rewrite 20y^{2}+y-1 as \left(20y^{2}-4y\right)+\left(5y-1\right).

$$\left(20y^{2}-4y\right)+\left(5y-1\right)$$

Factor out 4y in 20y^{2}-4y.

$$4y\left(5y-1\right)+5y-1$$

Factor out common term 5y-1 by using distributive property.

$$\left(5y-1\right)\left(4y+1\right)$$

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.

$$20y^{2}+y-1=0$$

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.

$$y=\frac{-1±\sqrt{1^{2}-4\times 20\left(-1\right)}}{2\times 20}$$

Square 1.

$$y=\frac{-1±\sqrt{1-4\times 20\left(-1\right)}}{2\times 20}$$

Multiply -4 times 20.

$$y=\frac{-1±\sqrt{1-80\left(-1\right)}}{2\times 20}$$

Multiply -80 times -1.

$$y=\frac{-1±\sqrt{1+80}}{2\times 20}$$

Add 1 to 80.

$$y=\frac{-1±\sqrt{81}}{2\times 20}$$

Take the square root of 81.

$$y=\frac{-1±9}{2\times 20}$$

Multiply 2 times 20.

$$y=\frac{-1±9}{40}$$

Now solve the equation y=\frac{-1±9}{40} when ± is plus. Add -1 to 9.

$$y=\frac{8}{40}$$

Reduce the fraction \frac{8}{40} to lowest terms by extracting and canceling out 8.

$$y=\frac{1}{5}$$

Now solve the equation y=\frac{-1±9}{40} when ± is minus. Subtract 9 from -1.

$$y=-\frac{10}{40}$$

Reduce the fraction \frac{-10}{40} to lowest terms by extracting and canceling out 10.

$$y=-\frac{1}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{1}{5} for x_{1} and -\frac{1}{4} for x_{2}.

$$20y^{2}+y-1=20\left(y-\frac{1}{5}\right)\left(y-\left(-\frac{1}{4}\right)\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$20y^{2}+y-1=20\left(y-\frac{1}{5}\right)\left(y+\frac{1}{4}\right)$$

Subtract \frac{1}{5} from y by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$20y^{2}+y-1=20\times \frac{5y-1}{5}\left(y+\frac{1}{4}\right)$$

Add \frac{1}{4} to y by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$20y^{2}+y-1=20\times \frac{5y-1}{5}\times \frac{4y+1}{4}$$

Multiply \frac{5y-1}{5} times \frac{4y+1}{4} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$20y^{2}+y-1=20\times \frac{\left(5y-1\right)\left(4y+1\right)}{5\times 4}$$

Multiply 5 times 4.

$$20y^{2}+y-1=20\times \frac{\left(5y-1\right)\left(4y+1\right)}{20}$$

Cancel out 20, the greatest common factor in 20 and 20.

$$20y^{2}+y-1=\left(5y-1\right)\left(4y+1\right)$$

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 20

$$x ^ 2 +\frac{1}{20}x -\frac{1}{20} = 0$$

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

$$r + s = -\frac{1}{20} rs = -\frac{1}{20}$$

Two numbers r and s sum up to -\frac{1}{20} exactly when the average of the two numbers is \frac{1}{2}*-\frac{1}{20} = -\frac{1}{40}. 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>

$$r = -\frac{1}{40} - u s = -\frac{1}{40} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{1}{20}

$$(-\frac{1}{40} - u) (-\frac{1}{40} + u) = -\frac{1}{20}$$

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

$$\frac{1}{1600} - u^2 = -\frac{1}{20}$$

Simplify the expression by subtracting \frac{1}{1600} on both sides

$$-u^2 = -\frac{1}{20}-\frac{1}{1600} = -\frac{81}{1600}$$

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

$$u^2 = \frac{81}{1600} u = \pm\sqrt{\frac{81}{1600}} = \pm \frac{9}{40}$$

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

$$r =-\frac{1}{40} - \frac{9}{40} = -0.250 s = -\frac{1}{40} + \frac{9}{40} = 0.200$$

Evaluate

$$20y^{2}+y-1$$