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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).
What do you do with c squared and b squared to get a squared?
You add c squared and b squared together to get a squared. This is based on the Pythagorean theorem, which states that in a right triangle, the square of the hypotenuse (a) is equal to the sum of the squares of the other two sides (b and c).
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.
