Describe the solution(s) to the equation x^2 + 5x + 4 = 0 by just determining discriminant. Show your work. Huh? cx

Question

Describe the solution(s) to the equation x^2 + 5x + 4 = 0 by just determining discriminant. Show your work.  Huh? cx

solomonzelman · Answer

Do you know what a discriminant is? If so, then you also know that the general fom of this type of equations is    ax^2+bx+c=0

Can you plug in your numbers ?

kewlgeek555 · Answer

I know about the general form of the equation is, but I seemed to have forgotten what the discriminant is. :3

solomonzelman · Answer

The discriminant is   b^2-4ac  (the thing inside the square rot in the quadratic formula)

kewlgeek555 · Answer

Is the discriminant just $$b^2-4ac$$or is it the actual $$\sqrt{b^2-4ac}$$?

johnweldon1993 · Answer

Square root is not included...so the first one you posted @kewlgeek555

kewlgeek555 · Answer

Okay, thanks.  So I have to plug in the values, right?  That would get me...
$$5^s-4(1)(4)$$Right?

kewlgeek555 · Answer

Oops, I meant to put five to the power of two, not eight. cx

johnweldon1993 · Answer

Now...the question doesn't state to SAY the answers to the question...merely to "describe" them

The objective of the discriminant is to tell you how many, and of what type, of solutions there are to your polynomial

johnweldon1993 · Answer

Lol yeah fix that exponent and then you will be correct...

$$\large 5^2 - 4(1)(4) = \space ?$$

kewlgeek555 · Answer

Oh okay.  Thanks!  Is there another method other than the discriminant to solve equations like these? (ax^2 + bx + c = 0)?  Are there specific times when you need to us them?

johnweldon1993 · Answer

Not done yet :P

still need to solve the

$$\large 5^2 - 4(1)(4) = \space ?$$

johnweldon1993 · Answer

When you solve that...look at your result...

The results of the discriminant will show you how many, and of what type, of solutions you have for this polynomial...


If the result > 0    you will have 2 real roots
If the result = 0    you will have 1 real root
If the result < 0    you will have 2 complex roots

johnweldon1993 · Answer

Also note...the discriminant does NOT give you the solutions to the equation...just tells you about them

kewlgeek555 · Answer

$$5^2-4(1)(4)$$$$25-4(1)(4)$$$$25-4(4)$$$$25-16$$$$9$$Okay, so since the result is greater than zero, I have two real roots. c:

johnweldon1993 · Answer

That is correct...2 real roots

That will be your "description" of your solutions

johnweldon1993 · Answer

Now...to answer your other question...

is there a way to solve a polynomial? (ax^2 + bx + c)

Yes of course...the quadratic equation   *this WILL give you your solutions*

$$\large \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

kewlgeek555 · Answer

Okay, and is there another way?  Because it is asking me for two methods.  Isn't there the b over 2a method, or is that for something else?

johnweldon1993 · Answer

No, that would solve for the vertex of your parabola...

2 methods huh?

well...you can use that equation

Or....you can factor

a.k.a make

ax^2 + bx + c     look something like    (x + a)(x + b)

*note...the 'a' in the first equation does NOT = the 'a' in the second part*   it is just the form I am describing

kewlgeek555 · Answer

Oh okay, I understand. :D

kewlgeek555 · Answer

Thanks, nice to see you again! ;D

johnweldon1993 · Answer

Anytime!