Deyj D.

asked • 07/13/15

quadratic formula

1/14=1/7x2-1/2x

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Inactive Tutor answered • 07/14/15

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1/14 = 1/7 x2 - 1/2 x
1/7 x2 - 1/2 x - 1/14 =0
multiply through by 14
2x2 - 7x - 1= 0
 
as Mike C worked out we have 
a=2, b= -7, c=-1
x = (7 ±√((-7)2-4(2)(-1))/(2(2))=
 
Don't forget that the 7 outside the square root is also divided by 4
 
x = (7 ±√57)/4
 
(7+√57)/4= 3.64
(7-√57)/4=-0.14
 
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Inactive Tutor

The answers on this forum are far more varied than the problems.  In fact, the same or very similar problems seem to appear as students make similar progress thru a subject.  So, why do we find typos and calculation errors and take very different approaches to solutions and even have conflict regarding "the better" or "the best" or "the correct" solution??
 
My first question would be, "Why, oh why, do so many tutors present a solution which has not been checked?"
 
Now, (TMI) I have fat fingers and am recovering from medical problems that have introduced typos, so I appreciate that this forum has spell checking (and I often write answers in MS Word before pasting them here).  Any answer anyone writes (it does not matter whether or not they are Albert Einstein [like Mike writes in his profile], that answer is characteristically not as good as one that has been spell checked.
 
My quote:  "With good people, a better process generally produces a better product."
 
Now, characteristically, a calculated check of a math solution produces a better solution [note: from this forum, I can say that almost everyone -- especially including me and Andrew -- has posted errors.  Wyzant has an accurate forum history that you can read if you doubt this.]
 
So, why do so many tutors post answers that have not been checked?
 
p.s., using Excel, I quickly determined that 3.64 and -0.14 both produce 0.0192, which is close to 0.  However, 8.89 produces 94.8342 and 5.11 produces 15.4542, which is a good clue that the problem has an error, the process has an error, but the person is very nice for posting free answers to questions from struggling students (especially at what are obviously late-night hours!).
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07/14/15

Inactive Tutor

Correction to my editing error in the last paragraph:  "... clue that the answer has an error, ..."    [it's early morning; truly, I did not intend to illustrate my point]
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07/14/15

Inactive Tutor

Unless this problem asks for a decimal solution, I would leave it as the square root solution.  That answer is correct without rounding things off.  Just wanted to show how much of a difference it makes when we make a small error.
 
As David W points out, many of us, myself included, are sometimes guilty of posting an answer that contains an error.  We should all check our answers.
 
I meant no disrespect to Mike C.  Just want to ensure the student gets the correct answer.  
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07/14/15

Inactive Tutor answered • 07/14/15

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y+1/14=1/7x2-1/2x
y+(14/1)(1/14)=(14/1)(1/7)x2(14/1)(1/2)x
y+1=2x2-7x
y+1-1=2x2-7x-1
y=2x2-7x-1
 
Quadratic Formula:  y=(-b±(√(b2-4(a)(c))))/2(a)   
y=2x2=(a)-7x=(b)-1(c)
y=(-(-7)±(√((-7)2-4(2)(-1))))/2(2)
y=7±(√(49-(8)(-1)))/4
y=7±(√(49+8))/4
y=7±(√(57))/4
 
y=7+(√(57))/4
y=8.89
 
y=7-(√(57))/4
y=5.11
 
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Mike,
You dropped the 7 out of the parentheses when you got to y= (7±√57)/4
 
(7+√57)/4=3.64
(7-√57)/4=-0.14
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07/14/15

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