1/4 X 1/2 X 1/4

$\exponential{(x)}{2} - 4 x - i $

\left(x-\left(ii-\sqrt{v}\right)\right)\left(x-\left(\sqrt{5}+2\right)\right)

x^{2}-4x-1

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x^{ii}-4x-1=0

Quadratic polynomial can exist factored using the transformation ax^{2}+bx+c=a\left(ten-x_{1}\right)\left(10-x_{ii}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

ten=\frac{-\left(-four\right)±\sqrt{\left(-four\right)^{ii}-four\left(-1\right)}}{ii}

All equations of the class ax^{ii}+bx+c=0 can exist 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(-four\correct)±\sqrt{16-four\left(-1\correct)}}{2}

Square -4.

10=\frac{-\left(-4\right)±\sqrt{xvi+4}}{2}

Multiply -4 times -1.

x=\frac{-\left(-4\right)±\sqrt{20}}{ii}

Add 16 to iv.

x=\frac{-\left(-4\right)±ii\sqrt{5}}{2}

Take the square root of 20.

10=\frac{iv±2\sqrt{five}}{2}

The opposite of -four is iv.

x=\frac{2\sqrt{five}+4}{2}

Now solve the equation x=\frac{4±two\sqrt{5}}{two} when ± is plus. Add iv to ii\sqrt{5}.

ten=\sqrt{5}+2

Carve up 4+2\sqrt{5} by 2.

x=\frac{4-two\sqrt{5}}{ii}

Now solve the equation x=\frac{4±2\sqrt{five}}{2} when ± is minus. Subtract 2\sqrt{five} from iv.

10=two-\sqrt{v}

Split 4-two\sqrt{5} by two.

ten^{2}-4x-1=\left(x-\left(\sqrt{5}+two\right)\correct)\left(x-\left(2-\sqrt{5}\right)\right)

Factor the original expression using ax^{2}+bx+c=a\left(10-x_{1}\right)\left(x-x_{2}\correct). Substitute 2+\sqrt{5} for x_{1} and 2-\sqrt{5} for x_{2}.

ten ^ 2 -4x -1 = 0

Quadratic equations such as this one can exist solved by a new direct factoring method that does not require approximate work. To use the straight factoring method, the equation must be in the course x^2+Bx+C=0.

r + southward = 4 rs = -1

Let r and southward be the factors for the quadratic equation such that ten^2+Bx+C=(10−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

r = 2 - u due south = 2 + u

Ii numbers r and s sum up to 4 exactly when the average of the 2 numbers is \frac{1}{2}*4 = 2. You lot can also see that the midpoint of r and s corresponds to the centrality of symmetry of the parabola represented past the quadratic equation y=x^ii+Bx+C. The values of r and s are equidistant from the centre past an unknown quantity u. Express r and southward 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) = -i

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

iv - u^2 = -one

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

-u^2 = -ane-four = -5

Simplify the expression past subtracting four on both sides

u^ii = five u = \pm\sqrt{v} = \pm \sqrt{5}

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

r =2 - \sqrt{5} = -0.236 due south = 2 + \sqrt{5} = 4.236

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

1/4 X 1/2 X 1/4,

Source: https://mathsolver.microsoft.com/en/solve-problem/%7B%20x%20%20%7D%5E%7B%202%20%20%7D%20%20-4x-1

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