2 X 3/4 + 1/4

$\exponential{(x)}{2} - iv x - 1 $

\left(x-\left(2-\sqrt{5}\right)\right)\left(10-\left(\sqrt{5}+ii\right)\right)

x^{2}-4x-1

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

ten=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-1\right)}}{ii}

All equations of the form ax^{2}+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 add-on and one when it is subtraction.

ten=\frac{-\left(-four\correct)±\sqrt{16-4\left(-1\right)}}{2}

Foursquare -four.

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

Multiply -iv times -i.

10=\frac{-\left(-4\right)±\sqrt{20}}{2}

Add 16 to 4.

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

Accept the square root of 20.

x=\frac{4±two\sqrt{5}}{ii}

The contrary of -4 is iv.

x=\frac{2\sqrt{5}+4}{2}

Now solve the equation x=\frac{4±2\sqrt{5}}{2} when ± is plus. Add 4 to 2\sqrt{5}.

x=\sqrt{5}+two

Divide 4+2\sqrt{5} by 2.

ten=\frac{4-two\sqrt{5}}{2}

Now solve the equation 10=\frac{four±2\sqrt{5}}{ii} when ± is minus. Decrease 2\sqrt{5} from 4.

10=two-\sqrt{5}

Separate 4-2\sqrt{5} past ii.

10^{2}-4x-1=\left(x-\left(\sqrt{v}+2\right)\right)\left(ten-\left(2-\sqrt{5}\right)\correct)

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

x ^ 2 -4x -1 = 0

Quadratic equations such as this ane tin be solved past a new direct factoring method that does non require gauge piece of work. To use the straight factoring method, the equation must be in the course ten^ii+Bx+C=0.

r + s = iv rs = -1

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

r = ii - u south = 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. Yous can also see that the midpoint of r and southward corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=ten^ii+Bx+C. The values of r and south 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' fashion='width: 100%;max-width: 700px' /></div>

(2 - u) (2 + u) = -1

To solve for unknown quantity u, substitute these in the product equation rs = -1

iv - u^2 = -i

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

-u^2 = -1-4 = -v

Simplify the expression by subtracting 4 on both sides

u^2 = v u = \pm\sqrt{5} = \pm \sqrt{v}

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

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

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

2 X 3/4 + 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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