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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
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
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.
(−b±√b2−4ac)÷2a the square root of b2−4ac entirely.
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)
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.
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
b2-4ac.
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For a quadratic equation of the form y = ax^2 + bx + c where a, b and c are real and a is non-zero, b^2 - 4ac enables you to distinguish between the following three possible outcomes:b^2 - 4ac > 0 => the equation has two real roots which are different,b^2 - 4ac = 0 => the equation has two real root which are coincident,b^2 - 4ac < 0 => the equation has two roots which form a complex conjugate pair.
b2-4ac