Question

Can someone tell me what the discriminant is in this quadratic equation.

-x2 – 2x+5=0

-x2 – 2+5=0
x= (-b±√(b^2-4ac))/2a

Answer 1

if the polynomial is in the form \(ax^2+bx+c=0\) the discriminant is \(b^2-4ac\)

you have \(a=-1, b=-2, c=5\)
discriminant is \((-2)^2-4(-1)(5)=\)

Answer 2

Thank you Whpalmer4. I understood that the discriminant is the number of roots but I got lost in my equation solving it.

Thanks you! Here is your medal!

Answer 3

oh, it's not the number of roots! but the value will tell you something about the kinds of roots you will get...

Answer 4

discriminant > 0 means 2 real roots
discriminant = 0 means 2 identical real roots
discriminant < 0 means 2 complex roots

(this is true for a quadratic, I make no promises about the discriminant for higher order polynomials)

Answer 5

ok, then what is the value of the roots in this equations?

Answer 6

the value of the roots will be
\[x = \frac{-b\pm\sqrt{b^2-4ac}}{2a} = \frac{-(-2)\pm\sqrt{(-2)^2-4(-1)(5)}}{2(-1) }= \]

Answer 7

\[=\frac{-2\pm\sqrt{24}} {-2}= -1\pm\sqrt{6}\]

Answer 8

Ok, when I plugged in the discriminant I got 24 as the result; but the way that I solved my equation is differs a bit. hmmmm

Answer 9

It's okay if the way you solved it differs a bit. What's not okay is if you got different answers :-)

Answer 10

did you solve it by completing the square?

\[-x^2-2x+5=0\]\[x^2+2x=5\]\[(x+1)^2 = 5+1^2\]\[x+1=\pm\sqrt{6}\]\[x=-1\pm\sqrt{6}\]

Answer 11

Yes I got the square root of 8 over 2.

Answer 12

for the equation \(-x^2-2x+5=0\)?

Answer 13

Square root (8) =2.83

Answer 14

Let's check that out:

\[-(\frac{\sqrt{8}}{2})^2-2(\frac{\sqrt{8}}{2})+5=0\]\[-\frac{8}{4}-\sqrt{8}+5=0\]Uh....not going to end well...

Answer 15

Ok, that means that I made an error somewhere

Answer 16

Let's try mine:

\[(-1-\sqrt{6})^2 = (1-\sqrt{6}-\sqrt{6}+(\sqrt{6})^2) = 7+2\sqrt{6}\]
\[-(7+2\sqrt{6}) - 2(-1-\sqrt{6}) + 5 = 0\]\[-7-2\sqrt{6}+2+2\sqrt{6} + 5= 0\]\[-7 + 2 + 5 = 0\checkmark\]

Answer 17

How do I check to see if -7+2+5=0 is equal to my equation -2-2+5=0

Answer 18

Well, the point is that when I put my solution into the original equation, the polynomial evaluated to 0, just like it is supposed to: \(-x^2-2x+5 = 0\)

\[-2-2+5=0\]\[-4+5=0\]\[1=0\]That's not true, is it? So if that's the result of putting your solution into the original equation, you've made a mistake either in finding the solution, or in testing it. Or maybe both :-)

Answer 19

But any valid solution to the equation will give you a true statement if substituted into the original equation you solved.

Answer 20

You are right, I tried to take a shortcut to find the discriminant and did not list the original equation:

x^2 -3x + 5= 0

Answer 21

\(x^2\) or \(-x^2\)?
Without the - sign, you'll have complex numbers for your solution, are you expecting that?

Answer 22

Yes, that is why I ended with the sqrt.(8)/2 man! Can you tell me what my discriminant is for this equation?

Answer 23

is it -x^2 or +x^2? You say one thing here, and another in the PM

Answer 24

Sorry about the mix up, I was writing from my notes and wanted to make sure I wrote my equation out correctly for class! I am straight now!

Answer 25

but I'm not! which one is it?

Answer 26

My final result is (8) = 2.83 or 0 + (-1.42) = -1.42

I find that the parabola crosses the x-axis also known as the x-intercepts.

Answer 27

For WHICH EQUATION?!?

Answer 28

the original equation which is -x^2 -3x+5=0

Answer 29

After dinner I will watch some media on Quadratic formulas and hopefully that will help!

Answer 30

Okay, how are you solving your quadratic equations? By the formula, factoring, or completing the square?

Answer 31

Solutions to \[-x^2-3x+5=0\] do not have an 8 in them.

Answer 32

If using the formula:
\[a = -1, b = -3, c = 5\]\[x = \frac{-b\pm\sqrt{b^2-4ac}}{2a} = \frac{-(-3)\pm\sqrt{(-3)^2-4(-1)(5)}}{2(-1)} \]\[= \frac{3\pm\sqrt{9+20}}{-2} = \frac{1}{2}(-3\pm\sqrt{29}) \]

Answer 33

When we simplify o + sqrt: 0-4 (-1) (2) = 0+ sqrt 8/-2 = sqrt -8/2 = sqrt 8/2

Answer 34

Where, oh where, are you finding a \(\sqrt{8}\)? Can you show me your work?

Answer 35

Is there a way that I can e-mail my Word document so that you can see it in the quadratic formula?