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Proof of the quadratic formula
Reminder
The general form of a quadratic equation is: $$ax^{2}+bx+c=0$$
x: unknown
a, b, c: constants with a != 0.
The general solution of a quadratic equation is: $$x={\frac {b\pm {\sqrt {b^{2}4ac}}}{2a}}$$
The plusminus symbol "±" indicates that the quadratic equation has two solutions (one using '+', the other using ''): $$x_{1}={\frac {b+{\sqrt {b^{2}4ac}}}{2a}}\quad {\text{and}}\quad x_{2}={\frac {b{\sqrt {b^{2}4ac}}}{2a}}$$ These two solutions are called the roots of the quadratic equation.
Geometrically, these roots represent the x values at which any quadratic parabola crosses the xaxis. $$y = ax^{2} + bx + c \quad > \quad x_{1},x_{2} \quad when \quad y = 0$$
Proof
Our goal is to transform our equation to the following form: $$x² + 2xb + b²$$ As: $$x² + 2xb + b² = (x + b)²$$ With the idea to be able to apply the square root on x².
We start with the general form: $$ax^{2}+bx+c=0$$ We multiply each side by 4a in order to transform the first terms to the square (2ax)² $$4a^{2}x^{2}+4abx+4ac=0$$ We rearrange $$4a^{2}x^{2}+4abx=4ac$$
If we say X = 2ax (for reading purpose): $$4a^{2}x^{2}+4abx=X^{2} + 2Xb$$ b² is the only term missing to reach our form: x² + 2xb + b²,
So we add b² to both sides (It is also called completing the square) $$4a^{2}x^{2}+4abx+b^{2}=b^{2}4ac$$ We use our formula to transform the left part $$(2ax+b)^{2}=b^{2}4ac$$
Great, Now we may use the square root operator.
We take the square root of both sides $$2ax+b=\pm {\sqrt {b^{2}4ac}}$$ We isolate x $$2ax=b\pm {\sqrt {b^{2}4ac}}$$ $$x={\frac {b\pm {\sqrt {b^{2}4ac}}}{2a}}$$
Proof done!
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