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Topic: **Unsolvable math problems homework****Question:**
I know how to solve them but they just seem unsolvable to me....i've tried to figure the problems out...but i can't find the solution...it's unfoiling.
The problems:
28) 2t (squared) + 8t - 64= 0
29) 3a (squared) - 36a + 81= 0
if you don't know what's unfoiling is, it's when you find out the factors of the whole number (when multiplied) that equal the number with the variable after it (not the squared one...the middle number!) when the factors of the whole number are added together....I'm mainly just trying to figure out what the factors are...but these two problems just seem unsolvable to me...do you agree? and if you know the factors...please tell me! Thank you :)

July 16, 2019 / By Cherice

28) I assume you mean 2t² + 8t - 64 = 0 You can factor out a 2 and be left with t² + 4t - 32 = 0 --> (t + 8)(t - 4) = 0 --> t = 4, -8 29) I assume you mean 3a² - 36a + 81 = 0 You can factor out a 3 and be left with a² - 12a + 27 = 0 --> (a - 9)(a - 3) = 0 --> a = 3, 9

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1. Just translate English into Algebra: "r is twice the value of p" ==> r = 2p "p is 8 less than s" ==> p = s - 8 Since p = s - 8, then s = p + 8. You can get this by remembering "families of facts" from grade school, or using algebra by adding 8 to both sides of p = s - 8. So, since r = 2p and s = p + 8, then r + s = 2p + (p + 8) r + s = 3p + 8 2. This is an old brainteaser. Since the trains are heading at each other, the relative speed between them is 55 mph + 35 mph = 90 mph. At 90 mph, how long does it take to close the 337.5 mile gap between them? The straight-arrow approach is that after x hours, train A is a = 55*x miles from Romeoville. Train B is b = 337.5 - 35*x miles away from Romeoville. When the trains meet, a=b, or 55x = 337.5 - 35x 90x = 337.5 and then solve for x to find that you have the same answer as before. 3. Let x peanut butter jars bought. Mike spent 8*$2 on the 8 jelly jars and x*$4 on peanut butter. This totals: 16 + 4x = 28 Solve this for x. Remember units: x is measured in "jars" or "jars of peanut butter". 4. First, look at the times. Let t be the walking time to school. Since the round-trip time is 8 hours (just who spends 8 hours/day walking to/from school, anyway?!?!), the return time is (8-t) hours. But the return time (8-t) is 2 hours longer than t, so: 8 - t = t + 2 6 = 2t t = 3 It took 3 hours to school and (8-3) = 5 hours to return. Next, use (distance) = (speed)*(time) for each trip to make two equations: d = (5 mph)*(3 hours) .... distance traveled on first trip is 5*3 = 15 miles 15 = (x mph)*(5 hours) 15 = 5x x = 15/5 = 3 mph 5. Read carefully. The problem is asking for the cost per cup of coffee. In this problem d dollars worth of coffee makes c cups of brewed coffee. The cost per cup is d/c. It's *already* in terms of c and d. The most important strategy is to read the question carefully. Above are some strategies, mostly in the examples, of how to solve problems like these.

An easy way to factor a trinomial into two binomials is by making the leading coefficient 1. 28) 2(t^2 + 4t - 32) = 0 Factor t^2 + 4t - 32: 2(t + 8)(t - 4) = 0 29) 3(a^2 - 12a + 27) 3(a - 9)(a - 3) = 0 You should know where to go from those spots. Hope I helped!

👍 40 | 👎 7

2(t^2 + 4t - 32) = 0 factor 1(-32) = -32 = -4*8 and -4 + 8 = 4 2(t^2 - 4t + 8t - 32) = 0 2[t(t - 4) + 8(t - 4)] = 0 2(t + 8)(t - 4) = 0 t = -8, 4 3(a^2 - 12a + 27) = 0 factor 1*27 = 27 = -9(-3) and -9 - 3 = -12 3(a^2 - 3a - 9a + 27) = 0 3[a(a - 3) - 9(a - 3)] = 0 3(a - 9)(a - 3) = 0 a = 9, 3

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I don't know what the t means in the first expression Either way, you shouldn't be asking other people to solve your problems for you; it's not a very efficient way of living your life (not to be rude). My advice is to consider a tutor.

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28) (2t - 8)(t + 8) = 0 2t - 8 = 0 t = 4 t + 8 = 0 t = -8 Answer: 4, -8 29) (3a - 9)(a - 9) 3a - 9 = 0 a = 3 a - 9 = 0 a = 9 Answer: 3, 9

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The reason teachers give you is that you need to be able to show how you got your answer. At higher math levels and professionally, this is not necessary. I agree with you. Much of what passes for math nowadays is busywork.

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