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Mathematics 16 Online
OpenStudy (anonymous):

Can someone help me:(? Decide which part of the quadratic formula tells you whether the quadratic equation can be solved by factoring. -b b2 - 4ac 2a Use the part of the quadratic formula that you chose above and find its value given the following quadratic equation: 4x2 + 6x + 2 = 0 (2 points)

OpenStudy (anonymous):

this was another one i dint quite get

pooja195 (pooja195):

The "discriminant" gives this information. It is the square root part of the quadratic formula. \[\sqrt{b^2-4ac}\]

pooja195 (pooja195):

From here solve the equation

pooja195 (pooja195):

Start by listing your abc values a=? b=? c=?

pooja195 (pooja195):

\[\huge~\rm~ax^2+bx+c=0\]

OpenStudy (anonymous):

4x^2+6x++2=0

OpenStudy (anonymous):

is that how it should be set up in order

pooja195 (pooja195):

yes but we need to figure out the values using \[\huge~\rm~ax^2+bx+c=0\] That equation can you list just the NUMBERS not the signs of variables

OpenStudy (anonymous):

im confused

OpenStudy (arindameducationusc):

@pooja195 after this, can you solve my question? http://openstudy.com/study#/updates/55b787e4e4b0f0c1f13fd1a9

pooja195 (pooja195):

Ok anything ^2 is the a value anything with an "x" is the b value anything with no varible is the c value now list the numbers :)

OpenStudy (anonymous):

so b^2 is A, -b is B, and C is 2a?

pooja195 (pooja195):

wut? o-o wow you really are confused ok lets start from scratch \[\huge~\rm~x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\] This is the quadratic formula \[\huge~\rm~ax^2+bx+c=0\] This is standered form The discernment is The "discriminant" gives this information. It is the square root part of the quadratic formula. \[\sqrt{b^2-4ac}\] We need to solve \[\huge~\rm~4x^2 + 6x + 2 = 0\] Notice how it all matches up with the standered form equation? \[\huge~\rm~ax^2+bx+c=0\]

OpenStudy (anonymous):

i know, i said that already

pooja195 (pooja195):

You have the numbers wrong a=4 b=6 c=2

OpenStudy (anonymous):

oh

OpenStudy (anonymous):

then why do we need just the numbers

pooja195 (pooja195):

We need just the numbers so that we can solve using the discernment

OpenStudy (anonymous):

ok

OpenStudy (arindameducationusc):

@pooja195 I request you... okay after this

pooja195 (pooja195):

\[\sqrt{b^2-4ac}\] So now we have our abc values \[\huge~\rm~ \sqrt{6^2-4(4)(2)}\] Solve that

OpenStudy (anonymous):

6^2-8 under a square root idk what

OpenStudy (anonymous):

i started with parenthesis and then multiplied 4 by (8)

pooja195 (pooja195):

ok lets do this step by step whats 6^2?

OpenStudy (anonymous):

36

pooja195 (pooja195):

ok leave that aside whats -4 times 4 ?

OpenStudy (anonymous):

-16

pooja195 (pooja195):

Ok good -16 x 2=?

OpenStudy (anonymous):

-32

pooja195 (pooja195):

good whats 36-32?

OpenStudy (anonymous):

4

OpenStudy (anonymous):

The 4ac

OpenStudy (anonymous):

4ac decides whether you get complex number or not.

pooja195 (pooja195):

good now whats the \[\huge~\rm \sqrt4\]

OpenStudy (anonymous):

2

pooja195 (pooja195):

@saseal you are confusing the asker we are already done with the problem.

OpenStudy (anonymous):

oh ok

pooja195 (pooja195):

@kinserkara so whats your final answer?

OpenStudy (anonymous):

sry about that

OpenStudy (anonymous):

2a

pooja195 (pooja195):

its ok :) @kinserkara now 2a it would just be x=2

OpenStudy (anonymous):

so the final answer is x=2

pooja195 (pooja195):

yep

OpenStudy (anonymous):

ok thanks

pooja195 (pooja195):

No problem :)

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