X 3 4x 1 0

$\exponential{(x)}{two} - four x + 1 = 0 $

ten=\sqrt{3}+two\approx iii.732050808

ten=2-\sqrt{3}\approx 0.267949192

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ten^{two}-4x+1=0

All equations of the form ax^{two}+bx+c=0 tin can exist solved using the quadratic formula: \frac{-b±\sqrt{b^{ii}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is improver and one when it is subtraction.

x=\frac{-\left(-4\right)±\sqrt{\left(-four\correct)^{2}-four}}{2}

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

x=\frac{-\left(-4\correct)±\sqrt{16-4}}{2}

Square -iv.

x=\frac{-\left(-4\right)±\sqrt{12}}{2}

Add 16 to -iv.

x=\frac{-\left(-4\right)±ii\sqrt{3}}{ii}

Take the foursquare root of 12.

x=\frac{4±2\sqrt{three}}{2}

The opposite of -four is 4.

x=\frac{2\sqrt{3}+iv}{2}

At present solve the equation x=\frac{4±2\sqrt{three}}{2} when ± is plus. Add 4 to 2\sqrt{3}.

ten=\sqrt{three}+2

Divide 4+ii\sqrt{3} by two.

10=\frac{4-two\sqrt{iii}}{ii}

Now solve the equation x=\frac{4±2\sqrt{3}}{2} when ± is minus. Decrease ii\sqrt{3} from 4.

x=ii-\sqrt{three}

Dissever four-2\sqrt{3} by 2.

x=\sqrt{3}+2 x=2-\sqrt{3}

The equation is now solved.

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

Quadratic equations such equally this one can exist solved by completing the foursquare. In order to consummate the square, the equation must first exist in the class ten^{2}+bx=c.

x^{ii}-4x+one-ane=-ane

Decrease 1 from both sides of the equation.

x^{2}-4x=-ane

Subtracting ane from itself leaves 0.

10^{2}-4x+\left(-2\right)^{2}=-1+\left(-two\right)^{2}

Split -4, the coefficient of the x term, by two to get -ii. And then add the square of -two to both sides of the equation. This stride makes the left hand side of the equation a perfect square.

x^{2}-4x+4=-i+iv

Foursquare -two.

10^{2}-4x+iv=3

Add -1 to 4.

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

Factor x^{2}-4x+4. In general, when x^{2}+bx+c is a perfect foursquare, it tin can always be factored as \left(x+\frac{b}{2}\right)^{two}.

\sqrt{\left(ten-ii\correct)^{two}}=\sqrt{3}

Accept the square root of both sides of the equation.

x-two=\sqrt{3} ten-ii=-\sqrt{three}

Simplify.

ten=\sqrt{3}+2 10=ii-\sqrt{iii}

Add together ii to both sides of the equation.

ten ^ ii -4x +one = 0

Quadratic equations such as this i can be solved by a new direct factoring method that does non crave judge work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.

r + southward = 4 rs = 1

Let r and s exist the factors for the quadratic equation such that x^2+Bx+C=(ten−r)(ten−southward) 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 upward to iv exactly when the average of the two numbers is \frac{one}{ii}*4 = 2. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented past the quadratic equation y=10^2+Bx+C. The values of r and south are equidistant from the center past an unknown quantity u. Limited 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) = 1

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

4 - u^ii = 1

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

-u^2 = 1-iv = -3

Simplify the expression by subtracting four on both sides

u^2 = 3 u = \pm\sqrt{3} = \pm \sqrt{3}

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

r =two - \sqrt{iii} = 0.2679491924311228 southward = 2 + \sqrt{iii} = 3.732050807568877

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

X 3 4x 1 0,

Source: https://mathsolver.microsoft.com/en/solve-problem/%7B%20x%20%20%7D%5E%7B%202%20%20%7D%20%20-4x%2B1%20%3D%20%200

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