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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
How do you find the discriminant of a polynomial?
The discriminant of the quadratic equation ax2+bx+c = 0 is the value of b2-4ac When b2-4ac = 0 then there are 2 equal roots. When b2-4ac > 0 then there are 2 different roots. When b2-4ac < 0 then there are no roots at all.
What is the value of b2-4ac equal to 0?
It is the value of the discriminant of a quadratic equation.
How do you tell if a discriminant is positive negative or zero?
With the standard notation, If b2 < 4ac then the discriminant is negative If b2 = 4ac then the discriminant is zero If b2 > 4ac then the discriminant is positive
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 is a x squared plus b x plus c equals 0?
x = (-b (+/-) root( b2 - 4ac)) / 2a Sorry about the messy answer, there are no square root symbols or plus-minus symbols. Here is the proof - ax2 + bx + c = 0 --------> multiply by 4a 4a2x2 + 4abx + 4ac = 0 4a2x2 + 4abx = -4ac ----------> add b2 to both sides 4a2x2 + 4abx + b2 = b2 - 4ac -----------> factorise LHS (2ax + b)2 = b2 - 4ac 2ax + b = (+/-) root( b2 - 4ac) 2ax = -b (+/-) root( b2 - 4ac) x = (-b (+/-) root( b2 - 4ac)) / 2a Maths works people.
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
What is the formula of the discriminant?
b2-4ac.
Write the quadratic equation in general form What is the value of b2 - 4ac?
-2
What is the value of x in ax2 bx c0?
ax2 + bx + c = 0 , find the value of x . b2-4ac>o x is real (2 different values will solve) b2-4ac=o -> a double root (a single real number will solve it) x=real numbers. b2-4ac<0 x= two complex number roots (either pure imaginary or a complex number with real and imaginary components)
Can there be a real and imaginary solution to a quadratic equation?
Yes, there can be a pure imaginary imaginary solution, as i2 =-1 and -i2 = 1. Or there can be a pure real solution or there can be a complex solution.For a quadratic equation ax2+ bx + c = 0, it depends on the value of the discriminant [b2 - 4ac], which is the value inside the radical of the quadratic formula.[b2 - 4ac] > 0 : Two distinct real solutions.[b2 - 4ac] = 0 : Two equal real solutions (double root).[b2 - 4ac] < 0 : Two complex solutions; they will be pure imaginary if b = 0, they will have both real and imaginary parts if b is nonzero.
How do you get from ax2 bx c to the quadratic formula?
ax2+bx+c=0 Multiply the whole equation by 4a: 4a2x2+4abx+4ac=0 Move the 4ac to the other side: 4a2b2+4abx=-4ac Add b2 to both sides: 4a2b2+4abx+b2=b2-4ac The left side of the equation is like (a+b)2=a2+2ab+b2, with a being 2ab and b being b: (2ax+b)2=b2-4ac Do a square root on both sides of the equation: 2ax+b=√(b2-4ac) Move the b to the other side of the equation: 2ax=-b±√(b2-4ac) Leave only x on the left side of the equation by dividing the right side by 2a: x=(-b±√(b2-4ac))/2a The previous explanation: aX2+bx+c is the same as x=-b (plus or minus the square root) of b2 - 4ac divided by two times a. x=(-b±√(b^2-4ac))/2a x can equal (-b+√(b^2-4ac))/2a ∆ (not delta) (-b-√(b^2-4ac))/2a
