Solving a disguised equation.
Solve x 2 − 5 x − 2 x 2 − 5 x + 3 = 1 2 . If the positive real root is in the form of a + c b , find ( 4 a + b + c )
The answer is 125.
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3 solutions
Sir you should show that why we can make this substitution else samuraiworm's solution is perfect
Great!!!! Did it in a similar ( not the same) mehtod!
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What was your method?
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jUST THE SAME. i had actually added three to both sides to make the expression more nice looking :P
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@Krishna Ar – hey [ @Krishna Ar ] You should definitely make a set on awesome equations!!!! and yeah, cool question :D
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@Krishna Ramesh – Thank you, :). I guess even you have the book Higher Algebra - right? Is that the secret to your good algebra @Krishna Ramesh . Have you taken up the NMTC? How did you fare?
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@Krishna Ar – yeah, I have higher algebra, ...no, I didnt take NMTC....I am solely concentrating on KVPY..
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@Krishna Ramesh – Ah! Well, How are you preparing for all those exams? @Krishna Ramesh
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@Krishna Ar – Here's a list of my resources: 1.Coaching 2.Books(like "Higher Algebra") 3.Brilliant.org!!!
@Krishna Ar – Who wrote this book?
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@Nathan Ramesh – I guess it was written by Knight...and someone else...you could google it perhaps :)... @Nathan Ramesh ....I would request that you tag me when you want to ask me a question because I dont really receive updates when someone rpelies within someone else's solution.
This problem took me way longer than it should have. I did it the same way as you I just didn't check for outliers so I thought there were two positive roots.
I also reached (5 + sqrt(113))/2 but I thought a,b, and must all be integers - _ -
It works when x = 6 if you use the negative value for the square root. 6 2 − 5 ( 6 ) + 3 = − 3
Can you do this?
I substituted
x
2
−
5
x
=
t
2
+
2
t
−
2
.
After solving,
t
=
±
4
.
Subtituting
t
in equation 1 and rest one can easily simplify.
Add 4 to both sides we get
x 2 − 5 x + 3 − 2 x 2 − 5 x + 3 + 1 = 1 6
Let a = x 2 − 5 x + 3 , a ≥ 0
( a − 1 ) 2 = 1 6
a = 5 ( a ≥ 0 > − 3 )
x 2 − 5 x + 3 = 2 5
x 2 − 5 x − 2 2 = 0
x = 2 5 + 1 1 3 ( 1 1 3 > 5 )
Me want harder. Me loving math. =3=3
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You can try this
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Me stuck till now T__T
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@Samuraiwarm Tsunayoshi – In what thing you are stuck
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@Ronak Agarwal – What you just gave me =="
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@Samuraiwarm Tsunayoshi – There no hidden identity involved in that question.Just use sine rule and a calculator.It is impossible to do it without a calculator.
I did in the same way
Should be +1=16
x 2 − 5 x − 2 x − 5 x + 3 = 1 2
x 2 − 5 x + 3 − 3 − 2 x − 5 x + 3 − 1 2 = 0
( x 2 − 5 x + 3 ) − 2 x − 5 x + 3 − 1 5 = 0
( x 2 − 5 x + 3 ) 2 − 2 x − 5 x + 3 − 1 5 = 0
Note that the above is an quadratic equation of x 2 − 5 x + 3 which factorized into the following:
( x 2 − 5 x + 3 + 3 ) ( x 2 − 5 x + 3 − 5 ) = 0
Since x 2 − 5 x + 3 = − 3 ⇒ x 2 − 5 x + 3 = 5
Therefore,
x 2 − 5 x + 3 = 2 5 ⇒ x 2 − 5 x − 2 2 = 0 ⇒ 2 5 ± 1 1 3
⇒ 4 a + b + c = 1 0 + 1 1 3 + 2 = 1 2 5
Let t = x 2 − 5 x , so the original equation now is: t − 2 t + 3 = 1 2 Move the root to one side: t − 1 2 = 2 t + 3 Square both sides and rearrange: t 2 − 2 8 t + 1 3 2 = 0 Factorize: ( t − 6 ) ( t − 2 2 ) = 0 The only solution for t is t = 2 2 because t = 6 does not satisfy the original equation.
Now, undo the change: 2 2 = x 2 − 5 x x 2 − 5 x − 2 2 = 0 Use the quadratic formula and find the positive root: x = 2 5 + 1 1 3 Here, a = 2 5 , b = 1 1 3 and c = 2 . Hence 4 a + b + c = 1 2 5 .