Here's the question you clicked on:
iloveboys
Solve: 0 = 6x^2 - 10x - 4.
you could factor that equation.
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
great answer man! provs
i have anoion... i have a graph liker this..
|dw:1337814199315:dw|the question is what quadratcic* equationis represtenesed below... than the graph
its somting l;ike that but more stright and more nice and even
answers: Non-factorable Trinomial Difference of Two Squares Not enough information Perfect Square Trinomial
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
okay thak u.. i was like whatttt.. lol
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?.
okay sory for all of this
no it's fine, just going to type it up elsewhere and copy/paste
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
let me know when you're ready for part 2
i am sorry i was try to understand it got it know go ahead when your ready
alright, no worries, take all the time you need and ask about anything
thank you i got it know.
alright, just checking
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
Solve: x2 - 2x - 24 = 0. Answer x = -4, x = 6 x = 4, x = -6 x = -4, x = -6 x = 4, x = 6
I already did this one, did you need to see the steps again?
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
I should be on tomorrow
okay well thank you again!
If I'm not on, you can email me or add me on msn or yahoo messenger either one works
you're welcome
sounds good.. what is your email?
jim_thompson5910@hotmail.com
ill porbley message you threw yahoo lol not msn lol
my messenger screen names are just jim_thompson5910
kkkk thank you byee can you repost theanser the question Solve: x2 - 2x - 24 = 0.
oh sry, one sec
dont be sorry i told u not to.. and than it asked me agiana nd i cant find it lol
lol alright, one sec while I type it up
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