anonymous
  • anonymous
Use the quadratic formula to solve 2y2 – 3y = 2. A. {½, 2} B. {½, –2} C. {–½, –2} D. {–½, 2}
Mathematics
schrodinger
  • schrodinger
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

pooja195
  • pooja195
Before the subtract the 2 so the the equation is = to 0 \[\huge 2y^2-3y-2\]
pooja195
  • pooja195
What are your a b c values?
pooja195
  • pooja195
Once you have those we can plug it into this equation AKA Quadratic Formula

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

pooja195
  • pooja195
\[\huge~x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]
anonymous
  • anonymous
I barley started to learn this, do you think you can walk me through it, step by step?
pooja195
  • pooja195
\[\huge~ax^2+bx+c\] \[\huge~2y^2-3y-2=0\] Anything that is ^2 will be the A just tell the number ignore the y^2 then anything with a plain vairble will just be the b value in this case it is -3 A number alone is the C value which in this case is -2
pooja195
  • pooja195
Got it so far?
anonymous
  • anonymous
yes
anonymous
  • anonymous
pooja195
  • pooja195
Good now we plug it in
KendrickLamar2014
  • KendrickLamar2014
Step 1. Subtract 2 from both sides: \[2y^2−3y−2=2−2\] \[= 2y^2−3y−2=0\]
anonymous
  • anonymous
KendrickLamar2014
  • KendrickLamar2014
Step 2. Factor the Left side: \[(2y+1)(y−2)=0\]
pooja195
  • pooja195
\[\huge~x=\frac{ -(-3)\pm \sqrt{(-3)^2-4(2)(-2)} }{ 2(2) } \]
KendrickLamar2014
  • KendrickLamar2014
Step 3. Set the factors to equal 0: \[2y+1=0~ or ~y−2=0\]
pooja195
  • pooja195
@KendrickLamar2014 it says to use the quadratic formula not factoring
KendrickLamar2014
  • KendrickLamar2014
I know, but you can get the same answer.
pooja195
  • pooja195
They need to use the quadratic formula its the directions.
KendrickLamar2014
  • KendrickLamar2014
\[= y = -\frac{ 1 }{ 2 }~or~ y = 2\]
pooja195
  • pooja195
Or just give them the answer...... @ninowletonzalez06 are you allowed to use factoring?
KendrickLamar2014
  • KendrickLamar2014
Yah, I was just showing her that there are many ways to solve this equation
pooja195
  • pooja195
Or have u not learned it because the Quadratic formula comes before factoring @KendrickLamar2014
pooja195
  • pooja195
Well now i bet that the asker is confused .-.
anonymous
  • anonymous
a liitle, can you explain it in the easiest form!
pooja195
  • pooja195
do you HAVE to use the quadratic formula?
anonymous
  • anonymous
by what the question is asking yes!
anonymous
  • anonymous
are you still there?
pooja195
  • pooja195
Ok :) All i did here was plug in the values
pooja195
  • pooja195
LaTex isnt working i cant finish it .-.
pooja195
  • pooja195
\[\huge~x=\frac{ -(-3) \pm \sqrt{(-3)^2-4(2)(-2)} }{ 2(2)} \]
anonymous
  • anonymous
okay!:)
pooja195
  • pooja195
\[\huge~x=\frac{ 3 \pm \sqrt{(-3)^2-4(2)(-2)} }{ 2(2)} \] simplify whats in the square root but DO NOT square root it
anonymous
  • anonymous
|dw:1433005023557:dw|
anonymous
  • anonymous
pooja195
  • pooja195
(-3)^2-4(2)(-2) thats all just miltiply :P
pooja195
  • pooja195
*multiply
anonymous
  • anonymous
9-(-16)
pooja195
  • pooja195
hmmm nope let me help u out :) i plugged it into a calc and got 25 \[\huge~x=\frac{ 3 \pm \sqrt{25} }{ 4} \]
anonymous
  • anonymous
oh okay!:)
pooja195
  • pooja195
Now we turn it into 2 diffrent problems \[\huge \frac{ 3+\sqrt{25} }{ 4 }\] and \[\huge \frac{ 3-\sqrt{25} }{ 4 }\]
pooja195
  • pooja195
Do you know how to solve?
anonymous
  • anonymous
3-5/4
anonymous
  • anonymous
pooja195
  • pooja195
im kinda confused on the solvng part :/ ive always used a clac to solve @KendrickLamar2014 do you know?
anonymous
  • anonymous
help!!!
pooja195
  • pooja195
@ybarrap can you help solve .-. please :)
KendrickLamar2014
  • KendrickLamar2014
What?
pooja195
  • pooja195
\(\color{blue}{\text{Originally Posted by}}\) @pooja195 Now we turn it into 2 diffrent problems \[\huge \frac{ 3+\sqrt{25} }{ 4 }\] and \[\huge \frac{ 3-\sqrt{25} }{ 4 }\] \(\color{blue}{\text{End of Quote}}\) this
pooja195
  • pooja195
i use a calc to solve that last step i dont know how to explain it
KendrickLamar2014
  • KendrickLamar2014
Oh, to explain or to solve it?
anonymous
  • anonymous
solve it
ybarrap
  • ybarrap
@pooja195 looks like you've got it covered here
KendrickLamar2014
  • KendrickLamar2014
I belive you take the square root of 25, then you add/subtract 3, lastly divide by 4
KendrickLamar2014
  • KendrickLamar2014
Ye, thats how you solve it ^
anonymous
  • anonymous
do u add or subtract
KendrickLamar2014
  • KendrickLamar2014
both
KendrickLamar2014
  • KendrickLamar2014
Its two equations
anonymous
  • anonymous
i confused
anonymous
  • anonymous
im**
KendrickLamar2014
  • KendrickLamar2014
\[\frac{ 3 + \sqrt25 }{ 4 }\] \[\frac{ 3-\sqrt25 }{ 4 }\] Two equations
KendrickLamar2014
  • KendrickLamar2014
\[\sqrt 25 = 5\] \[3+5 = 8\]
anonymous
  • anonymous
okay! i understand now!
KendrickLamar2014
  • KendrickLamar2014
\[\frac{ 8 }{ 4 } = 2\]
anonymous
  • anonymous
yes
KendrickLamar2014
  • KendrickLamar2014
Can you solve the other equation?
anonymous
  • anonymous
2/4=2
KendrickLamar2014
  • KendrickLamar2014
This equation:\[\frac{ 3-\sqrt25 }{ 4}\]
anonymous
  • anonymous
so -2?
KendrickLamar2014
  • KendrickLamar2014
No, lets go step-by-step. What is the square root of 25?
anonymous
  • anonymous
5
KendrickLamar2014
  • KendrickLamar2014
Yes, now what is 3-5?
anonymous
  • anonymous
-2
KendrickLamar2014
  • KendrickLamar2014
Yes, now what is \[\frac{ -2 }{ 4 }\]
anonymous
  • anonymous
-1/2
anonymous
  • anonymous
so the answer is D?
KendrickLamar2014
  • KendrickLamar2014
Yes
KendrickLamar2014
  • KendrickLamar2014
D is Correct

Looking for something else?

Not the answer you are looking for? Search for more explanations.