2x X 2 X 5
$\exponential{(x)}{2} – 2 x – 4 $
\left(ten-\left(1-\sqrt{v}\right)\right)\left(x-\left(\sqrt{v}+ane\right)\right)
ten^{2}-2x-four
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x^{2}-2x-4=0
Quadratic polynomial can be factored using the transformation ax^{two}+bx+c=a\left(x-x_{1}\correct)\left(ten-x_{ii}\correct), where x_{i} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.
x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-4\right)}}{ii}
All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{two}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when information technology is subtraction.
x=\frac{-\left(-two\right)±\sqrt{4-4\left(-iv\right)}}{2}
Foursquare -2.
x=\frac{-\left(-2\right)±\sqrt{4+16}}{two}
Multiply -four times -4.
10=\frac{-\left(-2\right)±\sqrt{xx}}{2}
Add four to sixteen.
ten=\frac{-\left(-2\correct)±2\sqrt{5}}{ii}
Take the square root of 20.
x=\frac{2±2\sqrt{v}}{2}
The opposite of -ii is 2.
ten=\frac{two\sqrt{v}+two}{ii}
Now solve the equation x=\frac{two±2\sqrt{5}}{two} when ± is plus. Add two to 2\sqrt{5}.
ten=\sqrt{5}+1
Divide two+2\sqrt{five} by 2.
x=\frac{2-2\sqrt{v}}{2}
At present solve the equation ten=\frac{2±2\sqrt{5}}{ii} when ± is minus. Subtract two\sqrt{5} from 2.
ten=i-\sqrt{5}
Separate 2-2\sqrt{5} by 2.
x^{2}-2x-4=\left(ten-\left(\sqrt{5}+1\right)\right)\left(x-\left(1-\sqrt{v}\right)\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(ten-x_{2}\right). Substitute 1+\sqrt{v} for x_{one} and 1-\sqrt{5} for x_{2}.
x ^ 2 -2x -four = 0
Quadratic equations such as this ane can be solved by a new direct factoring method that does titinada require gauge work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.
r + due south = 2 rs = -4
Let r and s exist the factors for the quadratic equation such that x^2+Bx+C=(x−r)(10−s) where sum of factors (r+s)=−B and the product of factors rs = C
r = i – u southward = 1 + u
Ii numbers r and s sum up to 2 exactly when the average of the two numbers is \frac{1}{two}*2 = 1. You lot can too run across that the midpoint of r and southward corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^ii+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 fashion='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' manner='width: 100%;max-width: 700px' /></div>
(ane – u) (1 + u) = -4
To solve for unknown quantity u, substitute these in the production equation rs = -4
ane – u^2 = -4
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^two = -4-1 = -5
Simplify the expression by subtracting ane on both sides
u^2 = 5 u = \pm\sqrt{5} = \pm \sqrt{v}
Simplify the expression by multiplying -one on both sides and have the square root to obtain the value of unknown variable u
r =1 – \sqrt{5} = -1.236 s = 1 + \sqrt{v} = 3.236
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
Source: https://mathsolver.microsoft.com/en/solve-problem/%7B%20x%20%20%7D%5E%7B%202%twenty%20%7D%20%20-2x-4
2x X 2 X 5,
Source: https://caribes.net/x-2-2x-4-0/
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