#1**+10 **

If we do this correctly....we will arrive at a familiar formula....let's see...

ax^2+bx+c=0 subtract c from both sides

ax^2 + bx = -c divide through by a

x^2 +(b/a)x = -c/a complete the square on x

x^2 + (b/a)x + [b^2] / [4a^2] = -c/a + [b^2] / [4a^2] factor the right side....get a common denominator on the left

[x + (b/2a)]^2 = [b^2 - 4ac] / [4a^2] take the square roots of both sides

x + (b/2a) = ±√[b^2 - 4ac]/[2a] subtract (b/2a) from both sides

x = (-b±√[b^2 - 4ac]) / [2a]

Voila!!!....we have the Quadratic Formula..... !!!!

CPhill
Mar 22, 2015

#1**+10 **

Best Answer

If we do this correctly....we will arrive at a familiar formula....let's see...

ax^2+bx+c=0 subtract c from both sides

ax^2 + bx = -c divide through by a

x^2 +(b/a)x = -c/a complete the square on x

x^2 + (b/a)x + [b^2] / [4a^2] = -c/a + [b^2] / [4a^2] factor the right side....get a common denominator on the left

[x + (b/2a)]^2 = [b^2 - 4ac] / [4a^2] take the square roots of both sides

x + (b/2a) = ±√[b^2 - 4ac]/[2a] subtract (b/2a) from both sides

x = (-b±√[b^2 - 4ac]) / [2a]

Voila!!!....we have the Quadratic Formula..... !!!!

CPhill
Mar 22, 2015