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Why is (4c to the power of 4)(ac to the power of 3)(3a to the power of 5) equal 12a to the power of 6 c to the power of 8?
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When the roots are equal of a quadratic equation?
Write the quadratic equation in the form ax2 + bx + c = 0 The roots are equal if and only if b2 - 4ac = 0. The expression, b2-4ac is called the [quadratic] discriminant.
What is the value of b2-4ac equal to 0?
It is the value of the discriminant of a quadratic equation.
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 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)
How would you describe the roots when b2-4ac is 0?
The roots will be equal
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.
When the roots are equal of a quadratic equation?
Write the quadratic equation in the form ax2 + bx + c = 0 The roots are equal if and only if b2 - 4ac = 0. The expression, b2-4ac is called the [quadratic] discriminant.
What is the value of b2-4ac equal to 0?
It is the value of the discriminant of a quadratic equation.
Why if the discriminant is zero is there no solution to the quadratic equation?
But there will be a solution if the discriminant is equal to zero: Real and different roots if b2-4ac > 0 Real and equal roots if b2-4ac = 0 But no real roots if b2-4ac < 0 in other words the graph wont make contact with or intercept the x axis.
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
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
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 is the quadratic formula?
x = [−b ± √(b2 − 4ac)]/2aA, B, and C can all correspond to the original quadratic equation as follows: ax2 + bx + c = 0The quadratic formula can only be used if the quadratic equation is equal to zero.If [ Ax2 + Bx + C = 0 ], thenx = [ -B +/- sqrt( B2 - 4AC ) ]/ 2A
