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Mathematics 80 Online
OpenStudy (sedatefrog712):

Medal if helped asap A quadratic equation is shown below: x2 + 5x + 4 = 0 Part A: Describe the solution(s) to the equation by just determining radicand. Show your work. (3 points) Part B: Solve 4x2 -12x + 5 = 0 using an appropriate method. Show the steps of your work, and explain why you chose the method used. (4 points) Part C: Solve 2x2 -10x + 3 = 0 by using a method different from the one you used in Part B. Show the steps of your work. (3 points)

OpenStudy (amistre64):

how do we define the radicand?

OpenStudy (studygurl14):

radicand, aka discriminant

OpenStudy (studygurl14):

discriminant: b^2 - 4ac

OpenStudy (amistre64):

lets see how much the asker already knows before hand

OpenStudy (studygurl14):

ok

OpenStudy (sedatefrog712):

i dont really remember much about the radicand

OpenStudy (amistre64):

the radicand is generally given as ... and study gurl has provided it

OpenStudy (studygurl14):

It is part of the quadratic formula: where \[ax^2+bx+c=0\] \[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

OpenStudy (amistre64):

this should be ringing some bells from what you have studied. so the next question to get your mind rolling would be what a,b, and c refer to

OpenStudy (sedatefrog712):

o yeah i remember how to use that formula

OpenStudy (studygurl14):

radicand is another name for square roots, cube roots, etc. (basically any type of root). It is specifying the stuff inside the square root

OpenStudy (amistre64):

i tend to prefer including the sqrt as part of the discriminant since it helps to see why the rules applied to it are as they are

OpenStudy (sedatefrog712):

ok

OpenStudy (studygurl14):

@SedateFrog712 Do you remember what the discriminant of a quadratic equation indicates about the solutions?

OpenStudy (studygurl14):

me too @amistre64

OpenStudy (anonymous):

you dont have to use that formula can solve by factoring as well

OpenStudy (amistre64):

KIO right, but the question is specific here.

OpenStudy (studygurl14):

Yes, you can @KlOwNlOvE , but the first question refers to the discriminant, which is part of the quadratic formula

OpenStudy (sedatefrog712):

tkind of @StudyGurl14

OpenStudy (anonymous):

ahh didnt read over that part my apologies

OpenStudy (amistre64):

good luck, youre in capable hands :)

OpenStudy (studygurl14):

If the discriminant is positive, there are two real roots. If it is zero, there is one real root (a double root). And if the discriminant is negative, there are two imaginary roots and no real roots

OpenStudy (sedatefrog712):

ok

OpenStudy (studygurl14):

So, do you know how to find the value of the discriminant? (Use the information given previously to help you!)

OpenStudy (sedatefrog712):

i would use the quadratic formula correct

OpenStudy (anonymous):

discriminant: b^2 - 4ac

OpenStudy (anonymous):

a=1 b=5 c=4

OpenStudy (sedatefrog712):

you mean a=2 @KlOwNlOvE

OpenStudy (studygurl14):

no, a =1

OpenStudy (sedatefrog712):

o ok

OpenStudy (studygurl14):

because it is x^2, so imagine the 1 in front of the x^2

OpenStudy (studygurl14):

If a=2, then it would be 2x^2

OpenStudy (studygurl14):

Okay, so if you have... \[b^2-4ac\] What does that equal?

OpenStudy (studygurl14):

(plug in the known values for a, b, and c)

OpenStudy (sedatefrog712):

5^2 - 4(1)(4) = 25 - 16= 9

OpenStudy (studygurl14):

Good job!

OpenStudy (studygurl14):

Okay, now refer back to what I said about discriminants. What does the 9 indicate about the solution of the equation?

OpenStudy (studygurl14):

"If the discriminant is positive, there are two real roots. If it is zero, there is one real root (a double root). And if the discriminant is negative, there are two imaginary roots and no real roots"

OpenStudy (sedatefrog712):

ok so 9 means there are two real roots

OpenStudy (sedatefrog712):

@StudyGurl14

OpenStudy (anonymous):

and sqrt of 9 is?

OpenStudy (sedatefrog712):

81

OpenStudy (anonymous):

no

OpenStudy (sedatefrog712):

3

OpenStudy (anonymous):

square root of 9 not 9 squared

OpenStudy (anonymous):

yes

OpenStudy (sedatefrog712):

uhh im stupid

OpenStudy (sedatefrog712):

ok so does that conclude a

OpenStudy (anonymous):

i think so solutions 3 3 ready for part b?

OpenStudy (sedatefrog712):

yep

OpenStudy (sedatefrog712):

@StudyGurl14 im ready for part b

OpenStudy (sedatefrog712):

i didnt

OpenStudy (studygurl14):

nvm sorry i didn't read far enough ahead

OpenStudy (studygurl14):

okay, so yeah, part b

OpenStudy (studygurl14):

you can do this three ways. Graphng, using the quadratic formula, or factoring. which method do you choose?

OpenStudy (anonymous):

would that be right? i haven't done this radicand things yet

OpenStudy (studygurl14):

would what be right?

OpenStudy (anonymous):

the solutions be 33 or just 3?

OpenStudy (studygurl14):

the solutions would be two real roots. you don't know them yet. The descriminant doesn't tell you what the solutions are, just how many and if they are real or imaginary

OpenStudy (studygurl14):

and the discriminant is 9, not 3

OpenStudy (sedatefrog712):

which method would be easiest

OpenStudy (studygurl14):

graphing is easiest if you have access to an online graphing calc like desmos.com. But for this one, I recommend factoring

OpenStudy (anonymous):

i find quadratic easier than the others

OpenStudy (sedatefrog712):

ok lets use factoring first

OpenStudy (anonymous):

oops\[4x^2-12x+5=0\]

OpenStudy (studygurl14):

Oh oops i was looking at the wrong one (sorry!) for this one, quadratic fwould be easier

OpenStudy (sedatefrog712):

ok let do quadratic

OpenStudy (studygurl14):

i was looking at x^2+5x + 4, sorry. i didn't realize part 2 had a different equation

OpenStudy (studygurl14):

Quadratic equation \[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

OpenStudy (studygurl14):

\[ax^2+bx+c=0\] \[4x^2-12x+5=0\] a= ? b = ? c= ?

OpenStudy (sedatefrog712):

a=4 b=12 c=5

OpenStudy (anonymous):

b is wrong

OpenStudy (studygurl14):

(don't forget the minus sign is included...)

OpenStudy (sedatefrog712):

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