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What is the value of b2-4ac equal to 0?
It is the value of the discriminant of a quadratic equation.
What is the quadratic formula of algebra?
(−b±√b2−4ac)÷2a the square root of b2−4ac entirely.
The Quadratic Formula may be used to solve any quadratic equation?
Yes. But note that if b2 - 4ac is negative, there are no real solutions to the quadratic equation to be found. When complex numbers are used, this is not a problem as sqrt(-1) = i and so if b2 - 4ac is negative, "sqrt(b2 - 4ac)" becomes "i sqrt(4ac - b2)", meaning the solutions are: x = -b/2a + i/2a sqrt(4ac-b2) x = -b/2a - i/2a sqrt(4ac-b2)
What determine how many solutions a quadratic equation will have and whether they are real or imaginary?
The quadratic equation is: Ax2+ Bx + C = 0-- The equation always has two solutions. They arex = 1/2A [ - B + sqrt(B2 - 4AC) ]x = 1/2A [ - B - sqrt(B2 - 4AC) ] .-- The solutions are real if ( B2 > or = 4AC ).-- The solutions are equal if ( B2 = 4AC ).-- The solutions are complex conjugates if ( B2 < 4AC ).-- The solutions are pure imaginary if ( B = 0 ) & (4AC>0 i.e. -4AC
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
