X 2 8x 20 0

$\exponential{(x)}{ii} - 8 x + 20 = 0 $

10=4+2i

10=iv-2i

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x^{2}-8x+20=0

All equations of the grade 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, ane when ± is addition and one when it is subtraction.

x=\frac{-\left(-viii\right)±\sqrt{\left(-8\correct)^{ii}-4\times 20}}{ii}

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -8 for b, and xx for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

x=\frac{-\left(-eight\right)±\sqrt{64-four\times twenty}}{2}

Square -8.

x=\frac{-\left(-8\right)±\sqrt{64-80}}{2}

Multiply -4 times 20.

10=\frac{-\left(-eight\right)±\sqrt{-16}}{2}

Add 64 to -fourscore.

10=\frac{-\left(-viii\right)±4i}{two}

Have the foursquare root of -16.

x=\frac{viii±4i}{2}

The opposite of -8 is eight.

x=\frac{eight+4i}{two}

At present solve the equation x=\frac{8±4i}{2} when ± is plus. Add eight to 4i.

ten=\frac{8-4i}{2}

Now solve the equation x=\frac{8±4i}{2} when ± is minus. Subtract 4i from 8.

10=4+2i 10=4-2i

The equation is now solved.

x^{2}-8x+xx=0

Quadratic equations such as this one tin can exist solved by completing the square. In lodge to consummate the square, the equation must showtime exist in the form x^{2}+bx=c.

x^{2}-8x+20-xx=-20

Subtract xx from both sides of the equation.

10^{2}-8x=-20

Subtracting 20 from itself leaves 0.

x^{2}-8x+\left(-four\right)^{2}=-20+\left(-four\right)^{ii}

Divide -8, the coefficient of the ten term, by 2 to get -4. And then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect foursquare.

x^{ii}-8x+16=-20+16

Foursquare -4.

10^{ii}-8x+sixteen=-4

Add -20 to 16.

\left(x-4\right)^{2}=-4

Factor x^{2}-8x+16. In general, when x^{two}+bx+c is a perfect square, it tin can always exist factored as \left(ten+\frac{b}{2}\right)^{two}.

\sqrt{\left(10-4\right)^{two}}=\sqrt{-four}

Take the square root of both sides of the equation.

x=4+2i x=4-2i

Add 4 to both sides of the equation.

x ^ ii -8x +20 = 0

Quadratic equations such as this one tin 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 course x^two+Bx+C=0.

r + s = 8 rs = 20

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

r = 4 - u south = 4 + u

Two numbers r and s sum upwards to 8 exactly when the average of the two numbers is \frac{1}{2}*8 = 4. You tin as well see that the midpoint of r and s 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 centre past an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.cyberspace/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

(4 - u) (iv + u) = 20

To solve for unknown quantity u, substitute these in the production equation rs = 20

16 - u^2 = twenty

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

-u^2 = 20-16 = 4

Simplify the expression by subtracting sixteen on both sides

u^2 = -four u = \pm\sqrt{-4} = \pm 2i

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

r =4 - 2i s = iv + 2i

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

X 2 8x 20 0,

Source: https://mathsolver.microsoft.com/en/solve-problem/%7B%20x%20%20%7D%5E%7B%202%20%20%7D%20%20-8x%2B20%3D0

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