“... mathematics is only the art of saying the same thing in different words” - B. Russell
The admins of this site are not happy if a full solution is given early, so I can only give hints.
In his reply Subhotosh Khan is assuming that this question is an exercise in using substitution and wants you to explain what you have learnt so far.
I believe that that there is a neater solution.
(1) j^2 + t = 10,050
(2) t^2 + j = 2,600
Make a new equation by subtracting terms of (1) from those of (2)
j^2 + t - t^2 - j = 10,050 - 2,600
see whether you can anything of this
I'm still not sure about how much one can give away immediately. I think that unless we know that the question in the OP was homework to practice substitution there's little point pursuing that path. Especially as it leads to to an awkward quartic.
In j^2 - t^2 + t - y = 10050 - 2600, I grouped the expressions on the left, saw "difference of two squares" so then could factorise with just two factors.
Saw which pair of values matched the factors of 10050 - 2600.
Must admit I assumed a whole number of dollars.
[tex](j^t + t^2) + t - j = 10050 - 2600 \implies (j - t)(j + t) - (j - t) = (j - t)(j + t - 1) = 7450.[/tex]
I agree that your factoring is valid and can be used to solve the problem by extensive trial and error.
But if you are going to assume whole numbers you can use the integer root theorem and get an answer very efficiently from the quartic.
[tex]t^4 - 5200t^2 + t + 6,749,950 = 0.[/tex] I have no idea why this does not render properly.
Factor the constant term with the absolutely obvious factor of 10
[tex]6,749,950 = 10 *674,995.[/tex]
Factor again by the obvious factor of 5
[tex]6,749,950 = 10 * 5 * 134999.[/tex]
It is obvious virtually by inspection that 2, 5, 10, and 25 are too small. 25 squared is 625 which would make j almost 2000, and that squared is way bigger than 10050. So try t = 50.
[tex]50^4 - 5200 * 50^2 + 50 = 6,250,000 - 50 * 2500 + 50 =[/tex]
[tex]6,250,050 - 13,000,000 = -\ 6,749,950.[/tex]
50 works.
I'd be amazed if this problem was not found in a section that mentions either the rational or integer root theorems or both.
Last edited by JeffM; 03-04-2018 at 04:28 PM. Reason: LaTeX error
Mon ami, I do not disagree that applying the answer suggested by the rational root theorem into the original equations is even more efficient. That, however, does not teach a lot about the potential of the theorem.
I suspect someone who does not know the rational root theorem and its corollary of the integer root theorem is not quite up to Newton's method for finding the zeroes of a polynomial. But it's definitely a method. And of course there is always a graphing calculator.
It could be a problem to teach follies of Newton's method.
If you start at 51 - it converges to 51.96143
If you start at 50.9 - it converges to 50
“... mathematics is only the art of saying the same thing in different words” - B. Russell
I reduced the equation to
(j−t)(j+t−1) = 2 x 5 x 5 x 149
and spotted factors 50 x 149 "by inspection", it did not require "extensive trial and error".
At least as quick as all the steps above.
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