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4x^{2}-20x-5=0
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.
x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 4\left(-5\right)}}{2\times 4}
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(-20\right)±\sqrt{400-4\times 4\left(-5\right)}}{2\times 4}
Square -20.
x=\frac{-\left(-20\right)±\sqrt{400-16\left(-5\right)}}{2\times 4}
Multiply -4 times 4.
x=\frac{-\left(-20\right)±\sqrt{400+80}}{2\times 4}
Multiply -16 times -5.
x=\frac{-\left(-20\right)±\sqrt{480}}{2\times 4}
Add 400 to 80.
x=\frac{-\left(-20\right)±4\sqrt{30}}{2\times 4}
Take the square root of 480.
x=\frac{20±4\sqrt{30}}{2\times 4}
The opposite of -20 is 20.
x=\frac{20±4\sqrt{30}}{8}
Multiply 2 times 4.
x=\frac{4\sqrt{30}+20}{8}
Now solve the equation x=\frac{20±4\sqrt{30}}{8} when ± is plus. Add 20 to 4\sqrt{30}.
x=\frac{\sqrt{30}+5}{2}
Divide 20+4\sqrt{30} by 8.
x=\frac{20-4\sqrt{30}}{8}
Now solve the equation x=\frac{20±4\sqrt{30}}{8} when ± is minus. Subtract 4\sqrt{30} from 20.
x=\frac{5-\sqrt{30}}{2}
Divide 20-4\sqrt{30} by 8.
4x^{2}-20x-5=4\left(x-\frac{\sqrt{30}+5}{2}\right)\left(x-\frac{5-\sqrt{30}}{2}\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5+\sqrt{30}}{2} for x_{1} and \frac{5-\sqrt{30}}{2} for x_{2}.
x ^ 2 -5x -\frac{5}{4} = 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 4
r + s = 5 rs = -\frac{5}{4}
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{5}{2} - u s = \frac{5}{2} + u
Two numbers r and s sum up to 5 exactly when the average of the two numbers is \frac{1}{2}*5 = \frac{5}{2}. 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{5}{2} - u) (\frac{5}{2} + u) = -\frac{5}{4}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{5}{4}
\frac{25}{4} - u^2 = -\frac{5}{4}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{5}{4}-\frac{25}{4} = -\frac{15}{2}
Simplify the expression by subtracting \frac{25}{4} on both sides
u^2 = \frac{15}{2} u = \pm\sqrt{\frac{15}{2}} = \pm \frac{\sqrt{15}}{\sqrt{2}}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{5}{2} - \frac{\sqrt{15}}{\sqrt{2}} = -0.239 s = \frac{5}{2} + \frac{\sqrt{15}}{\sqrt{2}} = 5.239
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.