why is b(square)-4ac > or equal to 0?

Question

why does my teacher say so?
i thought it is only b(square)-4ac>0
as the rule states so.
i dun understand why b(square)-4ac > or equal(bigger and equal sign) to 0?

Answer 1

It's allowed to be zero because the square root function allows zero.

Answer 2

If you're talking about the discriminant of the quadratic formula, then it can be greater than, less than, or equal to zero. If it is greater than zero, then there will be 2 x intercepts of the function. If it is equal to zero, then the function's vertex will lie on the x axis, meaning it has one x intercept, and if it is less than zero, then there will be no x intercept. A way of seeing this is that if you set the equation equal to zero, which is what the quadratic formula solves for (ax^2+bx+c=0) then you are looking for the point when the y value is zero (the x intercept). In the formula (x=[-b+-sqrt(b^2-4ac)]/2a) you must take the square root of b^2-4ac, and if that number is negative, you will have an imaginary number, which would mean there could be no solution and therefore no x intercept. If b^2=4ac was zero, then you would be adding or subtracting zero, which would give you the same answer, whether you added or subtracted zero, meaning there is only one solution, and therefore 1 x intercept. If b^2-4ac is positive, then you would get two different, real solutions, and therefore 2 different, real x intercepts

Answer 3

b^2 - 4ac is called the "discriminant" Δ of a quadratic function f(x)=ax^2+bx+c, a function that graphs out as a parabola for non-zero values of a.

You are probably at a stage in Maths where you cannot do square roots of a negative number (eg. SQRT(-4))

So you would like the discriminant Δ to be >=0 for a formula that will help you find the ROOTS (or ZEROs) of a quadratic function (where the parabola intersects with the x-axis).

The values of x for which f(x) = 0 is given by

x = (-b +/- SQRT( b^2 − 4ac)) / 2a

You might recognize the b^2-4ac inside the SQRT() there and if you can't do square roots of a negative number, then you will want it to be 0 or positive.

SQRT(0) is 0.

If Δ = 0 then it means the parabola has a "double root" - the two roots are the same number (ie. it touches the x-axis)

If Δ > 0 then the parabola cuts the x-axis in 2 places (and hence has 2 roots or zeros).

If Δ < 0 then the parabola lies above or below the x-axis entirely and has no REAL roots (or zeros) as it does not touch the x-axis.

Have a look at the wiki page

Answer 4

b^2 - 4ac can be less than, greater than, or equal to zero, depending on the quadratic equation you are looking at. It's just that you can do different things in the different cases.
2x^2 - 4x + 3 = 0
In this one, b^2 -4ac = -8, less than zero, so that means you can't solve the equation in the real number system.

x^2 - 5x + 6 = 0
In this one b^2 - 4ac = 1, greater than zero. That means you can solve the equation, and you will get two different solutions.

x^2 - 6x + 9 = 0
In this one b^2 - 4ac = 0. That means you can solve the equation, but there is only one number that is a solution. Since the quadratic has two factors, and each factor gives us a solution, we usually say we have two equal solutions.

So you see, it's not impossible for b^2 - 4ac to be negative, it's just that when it's negative, you can't solve the quadratic.

Answer 5

If b^2-4ac > 0, then you will have two real roots
If b^2-4ac = 0 then you will have a double real root
If b^2-4ac < 0, then you will have two imaginary roots

So if you want real roots, b^2-4ac = or > 0.

Answer 6

because the formula used to find roots is -b +- sqrt(b^2-4ac)/2a
if it is equal to 0, the sqrt of 0 is still possibile, which is 0
whereas the sqrt of a negative number is not defined in real
if it is equal 0, the 2 roots answers will be the same, so you have a perfect square

Answer 7

b^2-4ac = also known as delta
if delta > 0 than the 2 solutions are different
if delta = 0 than there is one solution
if delta < 0 there are no real solutions

Answer 8

complete the square
ax^2 + bx + c =
a( x^2 + 2b/(2a) x + c/a) =
a[ (x+b/(2a) )^2 - (b^2/(4a^2) - (c/a) ] =
a[ (x + b/((2a) )^2 - (b^2 - 4 ac)/(4a^2) ]

if this was a( (x+b/(2a))^2 - d^2) you can factorise as
a( x + b/(2a) + d ) ( x + b/(2a) - d)

so when the discriminant is zero the two real roots are the same
when the discriminant is >0 you can take square root and there are two real different roots
when the discriminant is <0 there are two complex conjugate roots.

Answer 9

that is known as the discriminant, ask your teacher about the quadratic formula (or better yet do a search on it)
when the discriminant is >= 0 you will have real valued answers, because it is usually placed under the square root symbol, and the squareroot of a negative number is not a real number

Answer 10

well is about that :
if b(square)-4ac >0 then you have 2 anwser (x1 and x2)
if b(square)-4ac = you have only 1 anwser
and if b(square)-4ac <0 there is no anwser