6x^2 - 28x + 49. Can this be simplified?
回答 (10)
2009-09-22 9:47 pm
✔ 最佳答案
In fact, there's lots of ways to factorize this into linear components. For example:(3x + (1/√2)7i - 7) (2x - (1/3)7i√2 - (1/3)14) = 6x² - 28x + 49
Is it simpler? No, but at least it's expressed as a product of linear terms.
I'm not really helping you, am I?
2009-09-22 10:28 am
6x² - 28x + 49 = 0
x² - 7/3x = - 49/6 + (- 7/3)²
x² - 7/3x = - 49/6 + 21/9
x² - 7/3x = - 147/18 + 42/18
(x - 7/3)² = - 35/6
x - 7/3 = ± 35i/√6
Answer: x = ± (35i/√6) + 7/3—imaginary solution
x² - 7/3x = - 49/6 + (- 7/3)²
x² - 7/3x = - 49/6 + 21/9
x² - 7/3x = - 147/18 + 42/18
(x - 7/3)² = - 35/6
x - 7/3 = ± 35i/√6
Answer: x = ± (35i/√6) + 7/3—imaginary solution
2016-11-03 2:52 pm
properly initially cancel out any exponents...a y from the the two cancels out so which you have... -7xy/28x^5 then do your x's -7y/28x^4 now do your numbers what proportion circumstances does 7 circulate into 28? 4 circumstances -y/4x^4 looking on your instructor they might ask you to the two positioned genuine or backside as destructive and additionally write it so as that each and every physique numbers are up genuine. in my adventure my instructor might want... -(a million/4)yx^-4 i think of it somewhat is so undemanding because it is going.
2009-09-22 5:23 pm
You could complete the square.
6x^2 - 28x + 49
x^2 - (14/3)x + (49/6) --> half of 14/3 is 7/3
x^2 - (14/3)x + (49/9) + [(49/6) - (49/9)]
[x - (7/3)]^2 + {[(9-6)49]/54}
[x - (7/3)]^2 + [3*49/54]
[x - (7/3)]^2 + (49/18) which won't simplify any further
6x^2 - 28x + 49
x^2 - (14/3)x + (49/6) --> half of 14/3 is 7/3
x^2 - (14/3)x + (49/9) + [(49/6) - (49/9)]
[x - (7/3)]^2 + {[(9-6)49]/54}
[x - (7/3)]^2 + [3*49/54]
[x - (7/3)]^2 + (49/18) which won't simplify any further
2009-09-22 11:59 am
It cannot be simplified.
2009-09-22 10:38 am
Then it equals what?
To answer the question provide the information
To answer the question provide the information
2009-09-22 10:33 am
since 28^2 - 4x6x49 is negative,
there are no real roots
there are no real roots
2009-09-22 10:31 am
Dear it cant be simplified as b^2 - ac is negative-
2009-09-22 10:29 am
if it's not equal to zero, what is it equal to? heheh and the first thing you would do is divide everything by 6 to get the x^2 by itself. you can plug it into a graphing calculator and see when the function intercepts the x-axis.
2009-09-22 10:29 am
Only if you want to solve for X to get a whole number.
6 is only divisible by 1, 2, 3 and 6
28 is only divisible by 1, 2, 4, 7, 14 and 28
49 is only divisible by 1, 7, and 49
There is no common term to simplify. 1 will keep all values the same.
6 is only divisible by 1, 2, 3 and 6
28 is only divisible by 1, 2, 4, 7, 14 and 28
49 is only divisible by 1, 7, and 49
There is no common term to simplify. 1 will keep all values the same.
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