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

ax2+bx+c=0ax^2 + bx + c = 0

change the subject to xx.

#Algebra #QuadraticEquations #Sharky

Note by Sharky Kesa
6 years, 10 months ago

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Comments

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 x2+bax+ca=0x^2+\frac{b}{a}x+\frac{c}{a}=0Use the transformation x=yb2ax=y-\frac{b}{2a}(yb2a)2+ba(yb2a)+ca=0\left(y-\frac{b}{2a}\right)^2+\frac{b}{a}\left(y-\frac{b}{2a}\right)+\frac{c}{a}=0y2bay+b24a2+bayb22a2+ca=0y^2-\frac{b}{a}y+\frac{b^2}{4a^2}+\frac{b}{a}y-\frac{b^2}{2a^2}+\frac{c}{a}=0y2=b24ac4a2y^2=\frac{b^2-4ac}{4a^2}x+b2a=±b24ac2ax+\frac{b}{2a}=\frac{\pm\sqrt{b^2-4ac}}{2a}x=b±b24ac2ax=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

Shriram Lokhande - 6 years, 10 months ago

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

{x=yan1nanf(x)=anxn+an1xn1++a1x+a0,an0\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}

    f(x)=anyn+pn2yn2+pn3yn3++p1y+p0\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 - 6 years, 10 months ago

Rewrite: x2+2xb2a=ca.x^2+2x \cdot\dfrac{b}{2a}=-\dfrac{c}{a}. Add (b/2a)2(b/2a)^2 to both sides and complete square: x2+2xb2a+(b2a)2=(b2a)2ca     (x+b2a)2=b24a2ca=b24ac4a2.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+b2a=±b24ac2a     x=b±b2+4ac2a.x+\dfrac{b}{2a}=\dfrac{\pm\sqrt{b^2-4ac}}{2a}~\implies x=\dfrac{-b\pm\sqrt{b^2+4ac}}{2a}.

Jubayer Nirjhor - 6 years, 10 months ago

congratulation for 135 / 135

sara sharma - 6 years, 10 months ago

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I don't know the score yet. Hopefully, it is 135/135.

Sharky Kesa - 6 years, 10 months ago
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