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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
How do you solve for the variable b in 2a plus 2b equals c?
2a + 2b = c subtract 2a from both sides 2a - 2a + 2b = c - 2a 2b = c - 2a divide both sides by 2 (2/2)b = (c - 2a)/2 b = (c - 2a)/2 --------------------
A equals one-half h b plus c in parentheses solve for b?
:a = .5(hb+c) :2a = hb+c :2a−c = hb :(2a−c)/h = b
If a plus a equals b and b plus c equals a and a-c equals b what is c?
a= (+a) or a= (-) b= 2a b= 2a c= (-a) c= (+a)
Why is the equation of the line of symmetry x equals negative b over 2a?
The question refers to the equation of a parabola, that is, a quadratic equation of the form y = ax2 + bx + c. Suppose x1 = -b/2a - z and x2 = -b/2a + z for some real number z. Then y1 = a*(-b/2a - z)2 + b*(-b/2a - z) + c = b2/4a + bz + az2 - b2/2a - bz + c = b2/4a + az2 - b2/2a + c and y2 = (-b/2a + z)2 + b*(-b/2a + z) + c = b2/4a - bz + az2 - b2/2a + bz + c = b2/4a + az2 - b2/2a + c So y1 = y2 thus, if x is the same distance (z) either side of -b/2a, then the corresponding y values are the same. And that, is what a line of symmetry means.
What is the axis of symmetry formula?
-b/2a. i think.To show this, consider this equation:y = ax² + bx + cFactor out the a:y = a(x² + bx/a + c/a)Then, complete the squares to get:y = a(x² + bx/a + (b/(2a))² + c/a - (b/(2a))²)= a((x + (b/2a))² + c/a - (b/(2a))²)= a(x + (b/2a))² + c - b/(4a)By the vertex form:y = a(x - h)² + k where x = h is the axis of symmetry.So the general axis of symmetry for the quadratic equation is x = -b/(2a).
Where did the quadratic formula 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
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
How do you find the max and min in a quadratic equation?
A quadratic function is of the form: f(x) = ax2 + bx + c where a ≠0 If a > 0 then the quadratic has a minimum but its maximum, asymptotically, is +∞. If a < 0 then the quadratic has a maximum but its minimum, asymptotically, is -∞. The extremum (whichever it is) is attained when x = -b/2a. The extreme value is f(-b/2a) = a*(-b/2a)2 + b(-b/2a) + c = b2/4a - b2/2a + c = -b2/4a + c
How do you factor 8a to the 3 plus 4a to the 2b-2ab to the 2-b to the 3?
Your notation is confusing. It looks like: 8a3 + 4a2b - 2ab2 - b3. So if you look at it, notice 23 = 8 & 22 = 4, so take c = (2*a), and you have c3 + c2b - cb2 - b3, Now note that if c = b, the polynomial is zero, so (c-b) or (b-c) is a factor. So divide by (c-b) using long division: (c - b)(c2 + 2cb + b2) {the second polynomial is a perfect square: (c+b)2}, so (c-b)(c+b)2, then substitute c = 2a, and get (2a-b)(2a+b)2 finally, multiply it out to make sure you get the original polynomial.
How do you factorise 4ab-2ac?
2a(2b-c) :P
