err let me rephrase my previous qn. it should be why b(square)-4ac≥ 0?

Question

i thought there are only 3 rules
why my teacher say there anothe rule which is
b(square)-4ac≥ 0
i dun understand why cuase since b(square)-4ac>0 too
den isnt both the same??

Answer 1

Delta = b^2 - 4ac

If delta <0, then you dont have real roots

If Delta = 0, then you have a doppel root. You add or to substract 0. But its a real root

If Delta >0 then you have 2 different roots

Ana

Answer 2

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 equation.

Answer 3

hi, using:
b(square)-4ac≥ 0
= that b(square)-4ac can be 'greater than OR EQUAL" to 0.

saying b(square)-4ac>0
=that b(square)-4ac can be greater than 0.

since there is times when it will be equal to 0, we need to say greater than or equal. if we had just > then we would be leaving out the time when it is 0, where as ≥ adds the time when it is 0.

for example if you want everyone who is over 3 years old you would want
(everyone)>3

if you wanted everyone 3 and older then you would have
(everyone)≥3

you can see the ≥ adds the "lower bound".

back to our case, we want to include the case when it is 0 so we have the ≥ to include the 0 also.

Note if b(square)-4ac is less then 0, then when we try and square root there is no answer. (unless you go into complex numbers but that is degree level maths!!!)

Answer 4

My previous answer still stands.
b^2-4ac≥ 0 simply combines two statemnets:
b^2-4ac > 0 and b^2 -4ac = 0 into 1 statement b^2-4ac≥ 0.

Answer 5

To have a REAL solution you can't have a neg square root.
It would be a complex solution.
Think about what the sqrt graph looks like.
The domain and range are positive.

Answer 6

if you reread the answers from your last question, you will find they answer this one as well.