Quadratic equation?

Question

For a quadratic equation ax^2 +bx + c =0 when a,b,c are real numbers... a cannot equal zero, but b or c may be equal to zero...... is this true and why?

Answer 1

If ax²2 + bx + c = 0
Then x = [(-b) ±{√(b² - 4ac)}]/2a

It is true that b or c or both can be zero, and that a cannot.

Here's why.

In the real numbers there is a thing called zero that has the peculiar property that whenever you multiply it by any real number, you get zero. It's the only real number that has that property,

For every other real number you can give me, I can give you another real number that when multiplied by your number you get 1.
If you give me 7, I’ll give you 1/7.
If you give me 0.5, I’ll give you 2.
If you give me -1/3, I’ll give you -3.

The number I give you is called the “multiplicative inverse” of the number you gave me. Because not too many people know that term, and because “divide by” is easier to say than “multiply by its multiplicative inverse,”
If you give most people 7, most people would say, to get 1 you “divide by 7”.

But there is no multiplicative inverse of 0. Remember, no matter what you multiply it by, you get 0. You cannot multiply it by any real number and get 1. Since it has no multiplicative inverse, and since multiplication by a number’s multiplicative inverse is called “dividing by that number” people (who aren’t as advanced in math as you now) are going to tell you, “you can’t divide by zero.”

With that in mind, you have 2a in the denominator. You and I now know that you are not “dividing by 2a” but multiplying by the multiplicative inverse of 2a. if a = 0, then 2a = 0 (because 0 multiplied by anything is 0). And since, if a = 0 then 2a = 0, and 2a would not have a multiplicative inverse… or, for the unenlightened…. “you can’t divide by zero.”

But 0/a (really 0 multiplied by the multiplicative inverse of a) = 0 (as long as a is not 0). So you can have 0 in the numerator of a fraction, but not in the denominator.

Answer 2

Because if a equaled zero, the equation would not be quadratic, but linear or a constant. And think about the quadratic equation: at the very end you divide the whole thing by 2a--but if a where zero, 20=0 and you can't divide by zero.

Answer 3

The quadratic equation is (-b +/- sqrt (b^2 - 4ac) / 2a if I remember correctly. So, a cannot equal 0 because you would be dividing by 0 which is not possible. b or c can be 0 because they are only in the numerator and would not cause an impossible result.

Answer 4

true. why? in order to have a quadratic equation you must have the minimum of a square term, i.e. x^2, otherwise you would have a linear equation.
besides, bx +c are mere transformations of the graph. for example, y = x^2 + 5 is the graph of x^2 shifted up 5 units. sililarly, y = x^2 + 2x +1 is the graph of x^2 shifted one unit to the left.

Answer 5

As Ana points out, with no xsquared term it would not be quadratic.

Such an equation would be 'linear' and the graph would be a straight line.

Answer 6

go to www.coolmath.com

It explains it the way everyone can understand.