Solve for x
x=-4
x=\frac{4}{5}=0.8
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a+b=16 ab=5\left(-16\right)=-80
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5x^{2}+ax+bx-16. To find a and b, set up a system to be solved.
-1,80 -2,40 -4,20 -5,16 -8,10
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 -80.
-1+80=79 -2+40=38 -4+20=16 -5+16=11 -8+10=2
Calculate the sum for each pair.
a=-4 b=20
The solution is the pair that gives sum 16.
\left(5x^{2}-4x\right)+\left(20x-16\right)
Rewrite 5x^{2}+16x-16 as \left(5x^{2}-4x\right)+\left(20x-16\right).
x\left(5x-4\right)+4\left(5x-4\right)
Factor out x in the first and 4 in the second group.
\left(5x-4\right)\left(x+4\right)
Factor out common term 5x-4 by using distributive property.
x=\frac{4}{5} x=-4
To find equation solutions, solve 5x-4=0 and x+4=0.
5x^{2}+16x-16=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{-16±\sqrt{16^{2}-4\times 5\left(-16\right)}}{2\times 5}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 16 for b, and -16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-16±\sqrt{256-4\times 5\left(-16\right)}}{2\times 5}
Square 16.
x=\frac{-16±\sqrt{256-20\left(-16\right)}}{2\times 5}
Multiply -4 times 5.
x=\frac{-16±\sqrt{256+320}}{2\times 5}
Multiply -20 times -16.
x=\frac{-16±\sqrt{576}}{2\times 5}
Add 256 to 320.
x=\frac{-16±24}{2\times 5}
Take the square root of 576.
x=\frac{-16±24}{10}
Multiply 2 times 5.
x=\frac{8}{10}
Now solve the equation x=\frac{-16±24}{10} when ± is plus. Add -16 to 24.
x=\frac{4}{5}
Reduce the fraction \frac{8}{10} to lowest terms by extracting and canceling out 2.
x=-\frac{40}{10}
Now solve the equation x=\frac{-16±24}{10} when ± is minus. Subtract 24 from -16.
x=-4
Divide -40 by 10.
x=\frac{4}{5} x=-4
The equation is now solved.
5x^{2}+16x-16=0
Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.
5x^{2}+16x-16-\left(-16\right)=-\left(-16\right)
Add 16 to both sides of the equation.
5x^{2}+16x=-\left(-16\right)
Subtracting -16 from itself leaves 0.
5x^{2}+16x=16
Subtract -16 from 0.
\frac{5x^{2}+16x}{5}=\frac{16}{5}
Divide both sides by 5.
x^{2}+\frac{16}{5}x=\frac{16}{5}
Dividing by 5 undoes the multiplication by 5.
x^{2}+\frac{16}{5}x+\left(\frac{8}{5}\right)^{2}=\frac{16}{5}+\left(\frac{8}{5}\right)^{2}
Divide \frac{16}{5}, the coefficient of the x term, by 2 to get \frac{8}{5}. Then add the square of \frac{8}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{16}{5}x+\frac{64}{25}=\frac{16}{5}+\frac{64}{25}
Square \frac{8}{5} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{16}{5}x+\frac{64}{25}=\frac{144}{25}
Add \frac{16}{5} to \frac{64}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{8}{5}\right)^{2}=\frac{144}{25}
Factor x^{2}+\frac{16}{5}x+\frac{64}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.
\sqrt{\left(x+\frac{8}{5}\right)^{2}}=\sqrt{\frac{144}{25}}
Take the square root of both sides of the equation.
x+\frac{8}{5}=\frac{12}{5} x+\frac{8}{5}=-\frac{12}{5}
Simplify.
x=\frac{4}{5} x=-4
Subtract \frac{8}{5} from both sides of the equation.
x ^ 2 +\frac{16}{5}x -\frac{16}{5} = 0
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 5
r + s = -\frac{16}{5} rs = -\frac{16}{5}
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 = -\frac{8}{5} - u s = -\frac{8}{5} + u
Two numbers r and s sum up to -\frac{16}{5} exactly when the average of the two numbers is \frac{1}{2}*-\frac{16}{5} = -\frac{8}{5}. 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>
(-\frac{8}{5} - u) (-\frac{8}{5} + u) = -\frac{16}{5}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{16}{5}
\frac{64}{25} - u^2 = -\frac{16}{5}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{16}{5}-\frac{64}{25} = -\frac{144}{25}
Simplify the expression by subtracting \frac{64}{25} on both sides
u^2 = \frac{144}{25} u = \pm\sqrt{\frac{144}{25}} = \pm \frac{12}{5}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{8}{5} - \frac{12}{5} = -4 s = -\frac{8}{5} + \frac{12}{5} = 0.800
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
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