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=== Completing the square by Śrīdhara's method === Instead of dividing by {{tmath|a}} to isolate {{tmath|\textstyle x^2\!}}, it can be slightly simpler to multiply by {{tmath|4a}} instead to produce {{tmath|\textstyle (2ax)^2\!}}, which allows us to complete the square without need for fractions. Then the steps of the derivation are:<ref name="Hoehn1975">{{citation |last=Hoehn |first=Larry |title=A More Elegant Method of Deriving the Quadratic Formula |journal=The Mathematics Teacher |year=1975 |volume=68 |issue=5 |pages=442–443 |jstor=27960212 |doi=10.5951/MT.68.5.0442 }}</ref> # Multiply each side by {{tmath|4a}}. # Add {{tmath|\textstyle b^2 - 4ac}} to both sides to complete the square. # Take the square root of both sides. # Isolate {{tmath|x}}. Applying this method to a generic quadratic equation with symbolic coefficients yields the quadratic formula: <math display=block>\begin{align} ax^2 + bx + c &= 0 \\[3mu] 4 a^2 x^2 + 4abx + 4ac &= 0 \\[3mu] 4 a^2 x^2 + 4abx + b^2 &= b^2 - 4ac \\[3mu] (2ax + b)^2 &= b^2 - 4ac \\[3mu] 2ax + b &= \pm \sqrt{b^2 - 4ac} \\[5mu] x &= \dfrac{-b\pm\sqrt{b^2 - 4ac }}{2a}. \vphantom\bigg) \end{align}</math> This method for completing the square is ancient and was known to the 8th–9th century Indian mathematician [[Śrīdhara]].<ref>Starting from a quadratic equation of the form {{tmath|1=\textstyle ax^2 + bx = c}}, Śrīdhara's method, as quoted by [[ Bhāskara II]] (c. 1150): "Multiply both sides of the equation by a number equal to four times the [coefficient of the] square, and add to them a number equal to the square of the original [coefficient of the] unknown quantity. [Then extract the root.]". {{pb}} {{harvnb|Smith|1923|loc=[https://archive.org/details/historyofmathema0001davi/page/446 {{pgs|446}}]}}</ref> Compared with the modern standard method for completing the square, this alternate method avoids fractions until the last step and hence does not require a rearrangement after step 3 to obtain a common denominator in the right side.<ref name=Hoehn1975/>
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