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Derive the quadratic formula. In other words, in the equation

\[ax^2 + bx + c = 0\]

change the subject to \(x\).

Note by Sharky Kesa
2 years, 7 months ago

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Quadratic formula can be derived more easily , we will use tschirnhaus transformation for the quadratic (used for reducing quartics and cubics)

The equation can be written as \[x^2+\frac{b}{a}x+\frac{c}{a}=0\]Use the transformation \(x=y-\frac{b}{2a}\)\[\left(y-\frac{b}{2a}\right)^2+\frac{b}{a}\left(y-\frac{b}{2a}\right)+\frac{c}{a}=0\]\[y^2-\frac{b}{a}y+\frac{b^2}{4a^2}+\frac{b}{a}y-\frac{b^2}{2a^2}+\frac{c}{a}=0\]\[y^2=\frac{b^2-4ac}{4a^2}\]\[x+\frac{b}{2a}=\frac{\pm\sqrt{b^2-4ac}}{2a}\]\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \] Shriram Lokhande · 2 years, 7 months ago

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@Shriram Lokhande This uses the interesting property (for those not too familiar with it)

\[\displaystyle \begin{cases}x=y-\frac{a_{n-1}}{n\cdot a_n} \\\\ f(x)=a_n x^n+a_{n-1}x^{n-1}+\cdots+a_1 x+a_0,\: a_n\neq 0\end{cases}\]

\[\displaystyle \implies f(x)=a_n y^n+p_{n-2}y^{n-2}+p_{n-3}y^{n-3}+\cdots +p_1 y+p_0\] Mathh Mathh · 2 years, 7 months ago

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Rewrite: \[x^2+2x \cdot\dfrac{b}{2a}=-\dfrac{c}{a}.\] Add \((b/2a)^2\) to both sides and complete square: \[x^2+2x \cdot\dfrac{b}{2a}+\left(\dfrac{b}{2a}\right)^2=\left(\dfrac{b}{2a}\right)^2-\dfrac{c}{a} ~\implies \left(x+\dfrac{b}{2a}\right)^2=\dfrac{b^2}{4a^2}-\dfrac{c}{a}=\dfrac{b^2-4ac}{4a^2}.\] Take square root and rearrange: \[x+\dfrac{b}{2a}=\dfrac{\pm\sqrt{b^2-4ac}}{2a}~\implies x=\dfrac{-b\pm\sqrt{b^2+4ac}}{2a}.\] Jubayer Nirjhor · 2 years, 7 months ago

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congratulation for 135 / 135 Sara Sharma · 2 years, 7 months ago

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@Sara Sharma I don't know the score yet. Hopefully, it is 135/135. Sharky Kesa · 2 years, 7 months ago

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