Solve 20y^2-9y-20

$$20 y ^ { 2 } - 9 y - 20$$

Factor

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

Solution Steps

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

$$a+b=-9 ab=20\left(-20\right)=-400$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -400.

$$1,-400 2,-200 4,-100 5,-80 8,-50 10,-40 16,-25 20,-20$$

Calculate the sum for each pair.

$$1-400=-399 2-200=-198 4-100=-96 5-80=-75 8-50=-42 10-40=-30 16-25=-9 20-20=0$$

The solution is the pair that gives sum -9.

$$a=-25 b=16$$

Rewrite 20y^{2}-9y-20 as \left(20y^{2}-25y\right)+\left(16y-20\right).

$$\left(20y^{2}-25y\right)+\left(16y-20\right)$$

Factor out 5y in the first and 4 in the second group.

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

Factor out common term 4y-5 by using distributive property.

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

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.

$$20y^{2}-9y-20=0$$

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.

$$y=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 20\left(-20\right)}}{2\times 20}$$

Square -9.

$$y=\frac{-\left(-9\right)±\sqrt{81-4\times 20\left(-20\right)}}{2\times 20}$$

Multiply -4 times 20.

$$y=\frac{-\left(-9\right)±\sqrt{81-80\left(-20\right)}}{2\times 20}$$

Multiply -80 times -20.

$$y=\frac{-\left(-9\right)±\sqrt{81+1600}}{2\times 20}$$

Add 81 to 1600.

$$y=\frac{-\left(-9\right)±\sqrt{1681}}{2\times 20}$$

Take the square root of 1681.

$$y=\frac{-\left(-9\right)±41}{2\times 20}$$

The opposite of -9 is 9.

$$y=\frac{9±41}{2\times 20}$$

Multiply 2 times 20.

$$y=\frac{9±41}{40}$$

Now solve the equation y=\frac{9±41}{40} when ± is plus. Add 9 to 41.

$$y=\frac{50}{40}$$

Reduce the fraction \frac{50}{40} to lowest terms by extracting and canceling out 10.

$$y=\frac{5}{4}$$

Now solve the equation y=\frac{9±41}{40} when ± is minus. Subtract 41 from 9.

$$y=-\frac{32}{40}$$

Reduce the fraction \frac{-32}{40} to lowest terms by extracting and canceling out 8.

$$y=-\frac{4}{5}$$

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

$$20y^{2}-9y-20=20\left(y-\frac{5}{4}\right)\left(y-\left(-\frac{4}{5}\right)\right)$$

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

$$20y^{2}-9y-20=20\left(y-\frac{5}{4}\right)\left(y+\frac{4}{5}\right)$$

Subtract \frac{5}{4} from y by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$20y^{2}-9y-20=20\times \frac{4y-5}{4}\left(y+\frac{4}{5}\right)$$

Add \frac{4}{5} to y by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$20y^{2}-9y-20=20\times \frac{4y-5}{4}\times \frac{5y+4}{5}$$

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

$$20y^{2}-9y-20=20\times \frac{\left(4y-5\right)\left(5y+4\right)}{4\times 5}$$

Multiply 4 times 5.

$$20y^{2}-9y-20=20\times \frac{\left(4y-5\right)\left(5y+4\right)}{20}$$

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

$$20y^{2}-9y-20=\left(4y-5\right)\left(5y+4\right)$$

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

$$x ^ 2 -\frac{9}{20}x -1 = 0$$

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

$$r + s = \frac{9}{20} rs = -1$$

Two numbers r and s sum up to \frac{9}{20} exactly when the average of the two numbers is \frac{1}{2}*\frac{9}{20} = \frac{9}{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>

$$r = \frac{9}{40} - u s = \frac{9}{40} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -1

$$(\frac{9}{40} - u) (\frac{9}{40} + u) = -1$$

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

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

Simplify the expression by subtracting \frac{81}{1600} on both sides

$$-u^2 = -1-\frac{81}{1600} = -\frac{1681}{1600}$$

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

$$u^2 = \frac{1681}{1600} u = \pm\sqrt{\frac{1681}{1600}} = \pm \frac{41}{40}$$

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

$$r =\frac{9}{40} - \frac{41}{40} = -0.800 s = \frac{9}{40} + \frac{41}{40} = 1.250$$

Evaluate

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