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anonymous
 4 years ago
Solve: 0 = 6x^2  10x  4.
anonymous
 4 years ago
Solve: 0 = 6x^2  10x  4.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you could factor that equation.

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1You 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0great answer man! provs

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i have anoion... i have a graph liker this..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1337814199315:dwthe question is what quadratcic* equationis represtenesed below... than the graph

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0its somting l;ike that but more stright and more nice and even

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0answers: Nonfactorable Trinomial Difference of Two Squares Not enough information Perfect Square Trinomial

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1since 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
 4 years ago
Best ResponseYou've already chosen the best response.0okay thak u.. i was like whatttt.. lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Part 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?.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okay sory for all of this

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1no it's fine, just going to type it up elsewhere and copy/paste

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1Part 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
 4 years ago
Best ResponseYou've already chosen the best response.1let me know when you're ready for part 2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i am sorry i was try to understand it got it know go ahead when your ready

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1alright, no worries, take all the time you need and ask about anything

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thank you i got it know.

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1alright, just checking

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1Part 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Solve: x2  2x  24 = 0. Answer x = 4, x = 6 x = 4, x = 6 x = 4, x = 6 x = 4, x = 6

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1I already did this one, did you need to see the steps again?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0im 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
 4 years ago
Best ResponseYou've already chosen the best response.1I should be on tomorrow

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okay well thank you again!

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1If I'm not on, you can email me or add me on msn or yahoo messenger either one works

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1you're welcome

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sounds good.. what is your email?

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1jim_thompson5910@hotmail.com

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ill porbley message you threw yahoo lol not msn lol

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1my messenger screen names are just jim_thompson5910

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0kkkk thank you byee can you repost theanser the question Solve: x2  2x  24 = 0.

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1oh sry, one sec

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dont be sorry i told u not to.. and than it asked me agiana nd i cant find it lol

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1lol alright, one sec while I type it up

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1Use 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
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