# Khan/khan-exercises

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 Solving quadratics by completing the square 1
randRangeNonZero( -10, 10 ) randRange( -5, 5 ) * 2 + ( X1 % 2 ) ( X1 + X2 ) * -1 X1 * X2 new Polynomial( 0, 2, [C, B, 1], "x" ) POLY.text() i18n._("or")

Complete the square to solve for x.

POLY_TEXT = 0

B/2 C * -1 + pow( B / 2, 2 ) X1 X2
B/2 C * -1 + pow( B / 2, 2 ) X2 X1
Completed Square:
(x + {} )^2 = {}

Solution:
X1

Completed Square:
(x + {}-X1 )^2 = {} 0

Solution:

Begin by moving the constant term to the right side of the equation.

x^2 + Bx = C * -1

We complete the square by taking half of the coefficient of our x term, squaring it, and adding it to both sides of the equation. Since the coefficient of our x term is B, half of it would be B / 2, and squaring it gives us \color{blue}{pow( B / 2, 2 )}.

x^2 + Bx \color{blue}{ + pow( B / 2, 2 )} = C * -1 \color{blue}{ + pow( B / 2, 2 )}

We can now rewrite the left side of the equation as a squared term.

( x + B / 2 )^2 = C * -1 + pow( B / 2, 2 )

The left side of the equation is already a perfect square trinomial. The coefficient of our x term is B, half of it is B / 2, and squaring it gives us \color{blue}{pow( B / 2, 2 )}, our constant term.

Thus, we can rewrite the left side of the equation as a squared term.

( x + B / 2 )^2 = C * -1 + pow( B / 2, 2 )

Take the square root of both sides.

x + B / 2 = \pmsqrt( C * -1 + pow( B / 2, 2 ) )

Isolate x to find the solution(s).

x = -B / 2\pmsqrt( C * -1 + pow( B / 2, 2 ) )

So the solutions are: x = -B / 2 + sqrt( C * -1 + pow( B / 2, 2 ) ) \text{ OR } x = -B / 2 - sqrt( C * -1 + pow( B / 2, 2 ) )

The solution is: x = -B / 2 + sqrt( C * -1 + pow( B / 2, 2 ) )

We already found the completed square: ( x + B / 2 )^2 = C * -1 + pow( B / 2, 2 )
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