Question

What is the value of the discriminant of 2x^2 + 4x – 2 = 0?
Answer

–20

32

0

16

Answer 1

all quadratic equations has this form:
\[ax^2+bx+c=0\]
to find discriminant you need to use this formula:
\[D=b^2-4ac\]
Now do it yourself?

Answer 2

Hi, so, let's try and answer this. I am going to show you something called the Quadratic Formula, that tells you when a quadratic equation is 0.

If the equation is of the form

ax^2 + bx + c = 0

Then the values of x that make the equation true are at

\[ {-b \pm \sqrt{b^2 - 4ac}} \over {2a} \]

That part under the square root is called the Descriminant.

Answer 3

So first let's try and figure out what 'a', 'b' and 'c' in your equation are, and then plug them into the formula for the discriminant. OK?

Answer 4

are you there?

Answer 5

yes shes there waiting for answer to write into test sheet :DD and not caring what discriminant is at all

Answer 6

trying to work it oout -________-

Answer 7

kis69fist23. The general equation is of the form
\[ ax^2 + bx + c = 0 \]
Your equation is
\[2x^2 + 4x - 2 = 0 \]
So from that, can you figure out what a, b, and c are?

Answer 8

so write down coefficients a,b,c out of that equation

Answer 9

What values of a, b, and c would make the general equation look like your equation?

Answer 10

Let's start with the x^2 term. What value of a would make "ax^2" look like "2x^2"?

Answer 11

Are you still there?

Answer 12

hey luv the answer is 16 :)

Answer 13

I guess she's gone.

Answer 14

would anyone be interested in knowing how to find the discriminant of an nth termed expression? I'll tell you... it's fun :)

Answer 15

sure, agreene. :)

Answer 16

Okay, so first... polynomial p(x) has a repeated root if and only if it shares a root with its derivative p'(x), the discriminant D(p) and the resultant R(p,p') both have the property that they vanish if and only if p has a repeated root, and they have almost the same degree and in fact are equal up to a factor of 1.

The resultant R(p,p') of the generalized polynomial:
\[p(x)=a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+...+a_2x^2+a_1x+a\]
is up to a factor of 1 equal to Sylvesters (2n-1)[cross](2n-1) determinate:
_
| a_n a_(n-1) ... a_1 a_0 0 ... ... 0
| 0 a_n a_(n-1) ... a_1 a_0 0 ... 0
| ...
| 0 ... 0 a_n a_(n-1)
| n*a_n
| 0
| ...
| 0
-
^^ yeah, never mind that I'll upload a photo from mathematica, thats really annoying to try and type out...

Anyway, from that we can then find the determinate D(p) of p(x) from:
\[D(p) = (-1)^{\frac12 n(n-1)}\frac1{a_n}R(p,p')\]

Answer 17

Here's that matrix I gave up on typing.

Answer 18

for instance, for the case:
\[p(x)= a_4x^4+a_3x^3+a_2x^2+a_1x+a_0\]
the determinate is as the attached picture shows.