2x 1 X 4 0

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

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x^{2}-4x+1=0

All equations of the form 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, one when ± is addition and i when information technology is subtraction.

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

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

x=\frac{-\left(-iv\right)±\sqrt{16-four}}{2}

Square -4.

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

Add together 16 to -4.

x=\frac{-\left(-4\right)±2\sqrt{3}}{ii}

Take the foursquare root of 12.

x=\frac{4±two\sqrt{3}}{2}

The contrary of -4 is 4.

x=\frac{2\sqrt{iii}+4}{ii}

Now solve the equation x=\frac{four±2\sqrt{3}}{2} when ± is plus. Add 4 to two\sqrt{3}.

10=\sqrt{three}+2

Carve up four+2\sqrt{iii} by ii.

x=\frac{4-2\sqrt{3}}{2}

Now solve the equation 10=\frac{4±ii\sqrt{3}}{2} when ± is minus. Decrease 2\sqrt{3} from four.

x=ii-\sqrt{3}

Dissever 4-two\sqrt{three} by 2.

ten=\sqrt{three}+2 ten=2-\sqrt{3}

The equation is now solved.

ten^{2}-4x+ane=0

Quadratic equations such equally this 1 can exist solved by completing the square. In society to complete the square, the equation must showtime be in the course x^{2}+bx=c.

x^{2}-4x+one-one=-1

Subtract 1 from both sides of the equation.

x^{2}-4x=-ane

Subtracting 1 from itself leaves 0.

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

Dissever -4, the coefficient of the x term, by 2 to become -2. Then add together the square of -2 to both sides of the equation. This step makes the left mitt side of the equation a perfect foursquare.

x^{2}-4x+4=-1+four

Square -2.

x^{2}-4x+4=iii

Add -i to 4.

\left(x-2\right)^{2}=3

Gene x^{2}-4x+4. In general, when 10^{ii}+bx+c is a perfect square, it can always be factored every bit \left(x+\frac{b}{2}\right)^{2}.

\sqrt{\left(x-2\right)^{two}}=\sqrt{3}

Take the foursquare root of both sides of the equation.

x-2=\sqrt{3} x-two=-\sqrt{three}

Simplify.

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

Add 2 to both sides of the equation.

x ^ 2 -4x +ane = 0

Quadratic equations such as this one can be solved by a new directly factoring method that does non require guess work. To apply the direct factoring method, the equation must be in the form x^2+Bx+C=0.

r + s = four rs = 1

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

r = 2 - u s = 2 + u

Ii numbers r and southward sum up to 4 exactly when the average of the two numbers is \frac{1}{two}*4 = two. You can also meet 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 center by an unknown quantity u. Express r and south 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) = one

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

4 - u^2 = 1

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

-u^2 = ane-4 = -three

Simplify the expression by subtracting iv on both sides

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

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

r =two - \sqrt{3} = 0.268 s = two + \sqrt{iii} = 3.732

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