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tim538Best ResponseYou've already chosen the best response.0
you could factor that equation.
 one year ago

DoomDudeBest ResponseYou've already chosen the best response.1
http://www.wolframalpha.com/input/?i=0+%3D+6x%5E2++10x++4
 one year ago

tim538Best ResponseYou've already chosen the best response.0
http://www.purplemath.com/modules/factquad.htm
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
You can factor, but let's use the quadratic equation to solve this \[\Large 0 = 6x^2  10x  4\] \[\Large 6x^2  10x  4 = 0\] \[\Large x = \frac{b\pm\sqrt{b^24ac}}{2a}\] \[\Large x = \frac{(10)\pm\sqrt{(10)^24(6)(4)}}{2(6)}\] \[\Large x = \frac{10\pm\sqrt{100(96)}}{12}\] \[\Large x = \frac{10\pm\sqrt{196}}{12}\] \[\Large x = \frac{10+\sqrt{196}}{12} \ \text{or} \ x = \frac{10\sqrt{196}}{12}\] \[\Large x = \frac{10+14}{12} \ \text{or} \ x = \frac{1014}{12}\] \[\Large x = \frac{24}{12} \ \text{or} \ x = \frac{4}{12}\] \[\Large x = 2 \ \text{or} \ x = \frac{1}{3}\] So you are correct iloveboys, nice work
 one year ago

DoomDudeBest ResponseYou've already chosen the best response.1
great answer man! provs
 one year ago

iloveboysBest ResponseYou've already chosen the best response.1
i have anoion... i have a graph liker this..
 one year ago

iloveboysBest ResponseYou've already chosen the best response.1
dw:1337814199315:dwthe question is what quadratcic* equationis represtenesed below... than the graph
 one year ago

iloveboysBest ResponseYou've already chosen the best response.1
its somting l;ike that but more stright and more nice and even
 one year ago

iloveboysBest ResponseYou've already chosen the best response.1
answers: Nonfactorable Trinomial Difference of Two Squares Not enough information Perfect Square Trinomial
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
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
 one year ago

iloveboysBest ResponseYou've already chosen the best response.1
okay thak u.. i was like whatttt.. lol
 one year ago

iloveboysBest ResponseYou've already chosen the best response.1
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?.
 one year ago

iloveboysBest ResponseYou've already chosen the best response.1
okay sory for all of this
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
no it's fine, just going to type it up elsewhere and copy/paste
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
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
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
let me know when you're ready for part 2
 one year ago

iloveboysBest ResponseYou've already chosen the best response.1
i am sorry i was try to understand it got it know go ahead when your ready
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
alright, no worries, take all the time you need and ask about anything
 one year ago

iloveboysBest ResponseYou've already chosen the best response.1
thank you i got it know.
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
alright, just checking
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
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 xaxis twice at 2 different spots y = x^2 + 2x + 1 only touches the xaxis at one spot only
 one year ago

iloveboysBest ResponseYou've already chosen the best response.1
Solve: x2  2x  24 = 0. Answer x = 4, x = 6 x = 4, x = 6 x = 4, x = 6 x = 4, x = 6
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
I already did this one, did you need to see the steps again?
 one year ago

iloveboysBest ResponseYou've already chosen the best response.1
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
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
I should be on tomorrow
 one year ago

iloveboysBest ResponseYou've already chosen the best response.1
okay well thank you again!
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
If I'm not on, you can email me or add me on msn or yahoo messenger either one works
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
you're welcome
 one year ago

iloveboysBest ResponseYou've already chosen the best response.1
sounds good.. what is your email?
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
jim_thompson5910@hotmail.com
 one year ago

iloveboysBest ResponseYou've already chosen the best response.1
ill porbley message you threw yahoo lol not msn lol
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
my messenger screen names are just jim_thompson5910
 one year ago

iloveboysBest ResponseYou've already chosen the best response.1
kkkk thank you byee can you repost theanser the question Solve: x2  2x  24 = 0.
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
oh sry, one sec
 one year ago

iloveboysBest ResponseYou've already chosen the best response.1
dont be sorry i told u not to.. and than it asked me agiana nd i cant find it lol
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
lol alright, one sec while I type it up
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
Use the quadratic formula to solve for x \[\Large x = \frac{b\pm\sqrt{b^24ac}}{2a}\] \[\Large x = \frac{(2)\pm\sqrt{(2)^24(1)(24)}}{2(1)}\] \[\Large x = \frac{2\pm\sqrt{4(96)}}{2}\] \[\Large x = \frac{2\pm\sqrt{100}}{2}\] \[\Large x = \frac{2+\sqrt{100}}{2} \ \text{or} \ x = \frac{2\sqrt{100}}{2}\] \[\Large x = \frac{2+10}{2} \ \text{or} \ x = \frac{210}{2}\] \[\Large x = \frac{12}{2} \ \text{or} \ x = \frac{8}{2}\] \[\Large x = 6 \ \text{or} \ x = 4\] So the answer is choice A
 one year ago
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