#+options: tex:t #+odt_math_syntax: starmath #+odt_extra_styles: #+odt_extra_styles: #+odt_extra_styles: #+begin_center *Finding Roots of a Quadratic Equation* #+end_center The process of /"completing the square"/ makes use of the algebraic identity #+NAME: step0 $$x^2 + 2hx + h^2 = (x+h)^2$$ which represents a well-defined algorithm that can be used to solve any quadratic equation. Starting with a quadratic equation in standard form, $ax^2 + bx + c = 0$ 1. Divide each side by $a$, the coefficient of the squared term. #+NAME: step1 $$x^2+{ b over a }x+{ c over a } = 0$$ 2. Subtract the constant term $c slash a$ from both sides. #+NAME: step2 $$x^2+{ b over a }x = -{ c over a }$$ 3. Add the square of one-half of $b slash a$, the coefficient of $x$, to both sides. This /"completes the square"/, converting the left side into a perfect square, as in [[step0]]. $$x^2 + 2({ b over { 2a } })x = -{ c over a }$$ #+NAME: step3 $$x^2 + 2({ b over { 2a } })x + ({ b over { 2a } })^2 = -{ c over a } + ({ b over { 2a } })^2$$ 4. Write the left side as a square and simplify the right side if necessary. #+NAME: step4 $$matrix { (x + b over { 2a })^2 # {} = {} # - { c over a } + { b^2 over { 4a^2 } } ## ~ # {} = {} # (-4ac + b^2) over { 4a^2 } }$$ 5. Produce two linear equations by equating the square root of the left side with the positive and negative square roots of the right side. #+NAME: step5 $$matrix{ (x + b over { 2a }) # {} = {}# +- sqrt{ { -4ac + b^2 } over { 4a^2 } } ## {} # {} = {} # +- { sqrt{ -4ac + b^2 } over { 2a } } }$$ 6. Solve each of the two linear equations. #+NAME: step6 $$matrix{ x # {} = {} # -b over { 2a } +- { sqrt{ -4ac + b^2 } over { 2a } } ## {} # {} = {} # { -b +- { sqrt{ -4ac + b^2 } } } over { 2a } }$$