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a+b=-4 ab=-5
To solve the equation, factor x^{2}-4x-5 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.
a=-5 b=1
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. The only such pair is the system solution.
\left(x-5\right)\left(x+1\right)
Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.
x=5 x=-1
To find equation solutions, solve x-5=0 and x+1=0.
a+b=-4 ab=1\left(-5\right)=-5
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-5. To find a and b, set up a system to be solved.
a=-5 b=1
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. The only such pair is the system solution.
\left(x^{2}-5x\right)+\left(x-5\right)
Rewrite x^{2}-4x-5 as \left(x^{2}-5x\right)+\left(x-5\right).
x\left(x-5\right)+x-5
Factor out x in x^{2}-5x.
\left(x-5\right)\left(x+1\right)
Factor out common term x-5 by using distributive property.
x=5 x=-1
To find equation solutions, solve x-5=0 and x+1=0.
x^{2}-4x-5=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(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-5\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -4 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-4\right)±\sqrt{16-4\left(-5\right)}}{2}
Square -4.
x=\frac{-\left(-4\right)±\sqrt{16+20}}{2}
Multiply -4 times -5.
x=\frac{-\left(-4\right)±\sqrt{36}}{2}
x=\frac{-\left(-4\right)±6}{2}
Take the square root of 36.
x=\frac{4±6}{2}
The opposite of -4 is 4.
x=\frac{10}{2}
Now solve the equation x=\frac{4±6}{2} when ± is plus. Add 4 to 6.
x=5
Divide 10 by 2.
x=\frac{-2}{2}
Now solve the equation x=\frac{4±6}{2} when ± is minus. Subtract 6 from 4.
x=-1
Divide -2 by 2.
x=5 x=-1
The equation is now solved.
x^{2}-4x-5=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.
x^{2}-4x-5-\left(-5\right)=-\left(-5\right)
Add 5 to both sides of the equation.
x^{2}-4x=-\left(-5\right)
Subtracting -5 from itself leaves 0.
x^{2}-4x=5
Subtract -5 from 0.
x^{2}-4x+\left(-2\right)^{2}=5+\left(-2\right)^{2}
Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-4x+4=5+4
Square -2.
x^{2}-4x+4=9
\left(x-2\right)^{2}=9
Factor x^{2}-4x+4. 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-2\right)^{2}}=\sqrt{9}
Take the square root of both sides of the equation.
x-2=3 x-2=-3
Simplify.
x=5 x=-1
Add 2 to both sides of the equation.
x ^ 2 -4x -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.
r + s = 4 rs = -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 = 2 - u s = 2 + u
Two numbers r and s sum up to 4 exactly when the average of the two numbers is \frac{1}{2}*4 = 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>
(2 - u) (2 + u) = -5
To solve for unknown quantity u, substitute these in the product equation rs = -5
4 - u^2 = -5
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -5-4 = -9
Simplify the expression by subtracting 4 on both sides
u^2 = 9 u = \pm\sqrt{9} = \pm 3
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =2 - 3 = -1 s = 2 + 3 = 5
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.