Solve: 0 = 6x^2 - 10x - 4.

- anonymous

Solve: 0 = 6x^2 - 10x - 4.

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- chestercat

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- anonymous

you could factor that equation.

- anonymous

how

- anonymous

http://www.wolframalpha.com/input/?i=0+%3D+6x%5E2+-+10x+-+4

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## More answers

- anonymous

http://www.purplemath.com/modules/factquad.htm

- anonymous

x+1/3, x=2???

- anonymous

x i meant to put x=

- anonymous

rewrite that please

- anonymous

x=-1/3, x=2????

- anonymous

is that correct?

- jim_thompson5910

You can factor, but let's use the quadratic equation to solve this
\[\Large 0 = 6x^2 - 10x - 4\]
\[\Large 6x^2 - 10x - 4 = 0\]
\[\Large x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
\[\Large x = \frac{-(-10)\pm\sqrt{(-10)^2-4(6)(-4)}}{2(6)}\]
\[\Large x = \frac{10\pm\sqrt{100-(-96)}}{12}\]
\[\Large x = \frac{10\pm\sqrt{196}}{12}\]
\[\Large x = \frac{10+\sqrt{196}}{12} \ \text{or} \ x = \frac{10-\sqrt{196}}{12}\]
\[\Large x = \frac{10+14}{12} \ \text{or} \ x = \frac{10-14}{12}\]
\[\Large x = \frac{24}{12} \ \text{or} \ x = \frac{-4}{12}\]
\[\Large x = 2 \ \text{or} \ x = -\frac{1}{3}\]
So you are correct iloveboys, nice work

- anonymous

thank you

- anonymous

great answer man! provs

- anonymous

i have anoion... i have a graph liker this..

- anonymous

|dw:1337814199315:dw|the question is what quadratcic* equationis represtenesed below... than the graph

- anonymous

its somting l;ike that but more stright and more nice and even

- anonymous

answers:
Non-factorable Trinomial
Difference of Two Squares
Not enough information
Perfect Square Trinomial

- jim_thompson5910

since the x intercepts are -2 and 2, we know that x = -2 and x = 2 are zeros
So...
x = -2 or x = 2
x + 2 = 0 or x - 2 = 0
(x + 2)(x - 2) = 0
x^2 - 4 = 0
Which is a difference of two squares since 4 is 2^2

- anonymous

okay thak u.. i was like whatttt.. lol

- anonymous

Part 1:
Solve each of the quadratic equations below and describe what the solution(s) represent to the graph of each. Show your work to receive full credit. •y = x2 + 3x + 2.
•y = x2 + 2x + 1.
Part 2:
Using complete sentences, answer the following questions about the two quadratic equations above.
•Do the two quadratic equations have anything in common? If so, what?.
•What makes y = x2 + 3x + 2 different from y = x2 + 2x + 1?.

- jim_thompson5910

one moment

- anonymous

okay sory for all of this

- jim_thompson5910

no it's fine, just going to type it up elsewhere and copy/paste

- jim_thompson5910

Part 1
y = x^2 + 3x + 2
0 = x^2 + 3x + 2
x^2 + 3x + 2 = 0
(x+2)(x+1) = 0
x+2=0 or x+1=0
x=-2 or x=-1
So the solutions to y = x^2 + 3x + 2 are x=-2 or x=-1
-------------------------------------------------------
y = x^2 + 2x + 1
0 = x^2 + 2x + 1
x^2 + 2x + 1 = 0
(x+1)^2 = 0
x+1 = 0
x = -1
So the only solution to y = x^2 + 2x + 1 is x = -1

- jim_thompson5910

let me know when you're ready for part 2

- anonymous

i am sorry i was try to understand it got it know go ahead when your ready

- jim_thompson5910

alright, no worries, take all the time you need and ask about anything

- anonymous

thank you i got it know.

- anonymous

now

- anonymous

go ahead

- jim_thompson5910

alright, just checking

- jim_thompson5910

Part 2
Similarities:
The two are similar in that they both are quadratic equations and graph parabolas. The two also open up in the same direction and have the same basic shape. The two quadratics are also factorable.
------------------------------------------------------------------
Differences:
y = x^2 + 3x + 2 has 2 solutions or zeros
y = x^2 + 2x + 1 has only 1 solution
y = x^2 + 3x + 2 intersects the x-axis twice at 2 different spots
y = x^2 + 2x + 1 only touches the x-axis at one spot only

- anonymous

Solve: x2 - 2x - 24 = 0.
Answer
x = -4, x = 6
x = 4, x = -6
x = -4, x = -6
x = 4, x = 6

- jim_thompson5910

I already did this one, did you need to see the steps again?

- anonymous

im done for today and im thankful that you helped me so much... hen will you be on gain. and no sorry i didnt post it

- jim_thompson5910

I should be on tomorrow

- anonymous

okay well thank you again!

- jim_thompson5910

If I'm not on, you can email me or add me on msn or yahoo messenger either one works

- jim_thompson5910

you're welcome

- anonymous

sounds good.. what is your email?

- jim_thompson5910

- anonymous

ill porbley message you threw yahoo lol not msn lol

- jim_thompson5910

my messenger screen names are just jim_thompson5910

- jim_thompson5910

so same name

- anonymous

kkkk thank you byee can you repost theanser the question Solve: x2 - 2x - 24 = 0.

- jim_thompson5910

oh sry, one sec

- anonymous

dont be sorry i told u not to.. and than it asked me agiana nd i cant find it lol

- jim_thompson5910

lol alright, one sec while I type it up

- anonymous

ok

- jim_thompson5910

Use the quadratic formula to solve for x
\[\Large x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
\[\Large x = \frac{-(-2)\pm\sqrt{(-2)^2-4(1)(-24)}}{2(1)}\]
\[\Large x = \frac{2\pm\sqrt{4-(-96)}}{2}\]
\[\Large x = \frac{2\pm\sqrt{100}}{2}\]
\[\Large x = \frac{2+\sqrt{100}}{2} \ \text{or} \ x = \frac{2-\sqrt{100}}{2}\]
\[\Large x = \frac{2+10}{2} \ \text{or} \ x = \frac{2-10}{2}\]
\[\Large x = \frac{12}{2} \ \text{or} \ x = \frac{-8}{2}\]
\[\Large x = 6 \ \text{or} \ x = -4\]
So the answer is choice A

- anonymous

thank you jim

- jim_thompson5910

anytime

- anonymous

bye

- jim_thompson5910

cya later

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