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# ax^2+bx+c=0 Solve for x.

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ax^2+bx+c=0 Solve for x.

Guest Mar 22, 2015

#1
+84237
+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
Sort:

#1
+84237
+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
#2
+91956
0

Thanks Chris :)

Melody  Mar 22, 2015

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