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Where did the quadratic equation come from?
The quadratic formula originated from the concept of completing the square. let's take ax2 + bx + c = 0. To complete the square, solve for x. Subtract c. ax2 + bx = -c. Then divide by a [notice- if there is no a value, then a=1]. x2 + bx/a = -c/a. Add (b/2a)2 to both sides. x2 + bx/a + b2/4a2 = -c/a + b2/4a2 Factor/Reformat. (x + b/2a)2 = (b2-4ac) / 4a2 (x + b/2a)2 = [(b2-4ac) / 2a]2 Square-root both sides. x + b/2a = ± √(b2-4ac) / 2a Subtract b/2a. x = -b/2a ± √(b2-4ac) / 2a Combine terms. x = [-b ± √(b2-4ac)] / 2a
How do you prove the quadratic formula?
The essence of the proof is simply to complete the square for a generalised quadratic equation. Like this:ax2 + bx + c = 0Take 'a' outside:a[x2 + bx/a + c/a] = 0Divide through by 'a':x2 + bx/a + c/a = 0Complete the square:(x + b/2a)2 - b2/4a2 + c/a = 0Rearrange to find x:(x + b/2a)2 = b2/4a2 - c/ax + b/2a = (+/-)sqrt[b2/4a2 - c/a]x = -b/2a (+/-) sqrt[b2/4a2 - c/a]Finally, fiddle around so that (1/2a) can be taken out as a common factor:x = -b/2a (+/-) sqrt[b2/4a2 - 4ac/4a2]x = -b/2a (+/-) sqrt[(1/4a2)(b2 - 4ac)]x = -b/2a (+/-) sqrt(1/4a2)sqrt(b2 - 4ac)x = -b/2a (+/-) (1/2a)sqrt(b2 - 4ac)x = [ -b (+/-) sqrt(b2 - 4ac) ] / 2a.
If a plus b equals c what does a plus c equal?
A+c= 2a+b
Ax2 plus bx plus c equals 0?
x=(-b+sqrt(b^2-4*a*c))/(2a) and (-b-sqrt(b^2-4*a*c))/(2a) (This is called the quadratic formula)
What Gpa is 2a's 2 b's and 2 c's?
3.0
