Question

Derive the quadratic formula from the standard form (ax2 + bx + c = 0) of a quadratic equation by following the steps below.

a.Divide all terms in the equation by a.
b.Subtract the constant (the term without an x) from both sides.
c.Add a constant (in terms of a and b) that will complete the square.
d.Take the square root of both sides of the equation.
e.Solve for x.

Answer 1

Everyone should do this derivation, because otherwise the "quadratic formula" is some sort of "magic" LOL...

ax^2+bx+c=0  

x^2+bx/a+c/a=0

x^2+bx/a=-c/a

x^2+bx/a+b^2x/(4a^2)=b^2/(4a^2)-c/a

(x+b/(2a))^2=(b^2-4ac)/(4a^2)

x+b/(2a)=±√(b^2-4ac)/(2a)

x=-b/(2a)±√(b^2-4ac)/(2a)

x=(-b±√b^2-4ac)/(2a)

Answer 2

Answer:

See below.

Step-by-step explanation:

We are going to take the quadratic formula ax²+bx+c=0

a.Divide all terms in the equation by a.

$\frac{ax^{2} }{a} +\frac{b}{a}x+\frac{c}{a}=0\\\\x^{2} +\frac{b}{a}x+\frac{c}{a}  =0$

b.Subtract the constant (the term without an x) from both sides.

$x^{2} +\frac{b}{a}x+\frac{c}{a} -\frac{c}{a}  =\frac{-c}{a} \\\\x^{2} +\frac{b}{a}x=-\frac{c}{a}$

c.Add a constant (in terms of a and b) that will complete the square.

$x^{2} +\frac{b}{a}x=-\frac{c}{a}\\\\[tex]x^{2} +\frac{b}{a}x+\frac{b^{2} }{4a^{2} }  =-\frac{c}{a} +\frac{b^{2} }{4a^{2}}\\\\(x+\frac{b}{2a}) ^{2}  =\frac{-4ac+b^{2} }{4a^{2} }$

d.Take the square root of both sides of the equation.

$x+\frac{b}{2a}  =[tex]\\x+\frac{b}{2a} =\frac{\sqrt{-4ac+b^{2} } }{2a}\\x=\frac{\sqrt{-4ac+b^{2} } }{2a}-\frac{b}{2a}$

e.Solve for x.

$x=-b+\frac{\sqrt{b^{2}-4ac } }{2a}$