# Thread: Quadratic Simultaneous Eqns: ""J and T share money. If I square J's money and add..."

1. Originally Posted by Jim77
Apparently so...

so (2600 - t^2)^2 + t = 10,050

Is this correct?

This is where I have no idea on how to solve for t...
(2600 - t^2)^2 + t = 10,050

Expand

t^4 - 5200* t^2 + t + 6749950 = 0

In a previous post you were asked:

What method(s) have you been taught for solving systems of non-linear equations?

2. Originally Posted by Jim77
Sorry could you expand on that a little bit please?
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

3. Originally Posted by Jonathan
(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
Yo Jonathan!
That'll simplify to: 2j = 1 +- SQRT(4t^2 - 4t + 29801)
So (I think!) you'll still end up with a 4th degree...
Or do I need another coffee

j^2 + t = 10,050
Btw, the value of j is "evident" from just "looking" at that equation.
Closest square < 10050

4. Originally Posted by Denis
Yo Jonathan!
That'll simplify to: 2j = 1 +- SQRT(4t^2 - 4t + 29801)
So (I think!) you'll still end up with a 4th degree...
Or do I need another coffee

j^2 + t = 10,050
Btw, the value of j is "evident" from just "looking" at that equation.
Closest square < 10050
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.

5. Originally Posted by Jonathan
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.
$(j^t + t^2) + t - j = 10050 - 2600 \implies (j - t)(j + t) - (j - t) = (j - t)(j + t - 1) = 7450.$

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.

$t^4 - 5200t^2 + t + 6,749,950 = 0.$ I have no idea why this does not render properly.

Factor the constant term with the absolutely obvious factor of 10

$6,749,950 = 10 *674,995.$

Factor again by the obvious factor of 5

$6,749,950 = 10 * 5 * 134999.$

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.

$50^4 - 5200 * 50^2 + 50 = 6,250,000 - 50 * 2500 + 50 =$

$6,250,050 - 13,000,000 = -\ 6,749,950.$

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.

6. j^2 + t = 10,050

j = sqrt(10000) = 100
t = 10050 - 10000 = 50

Ahem

7. Originally Posted by Denis
j^2 + t = 10,050

j = sqrt(10000) = 100
t = 10050 - 10000 = 50

Ahem
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.

8. Originally Posted by Subhotosh Khan
Or use Newton's method - converges in about 3 steps.
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.

9. 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

10. Originally Posted by JeffM
$(j^t + t^2) + t - j = 10050 - 2600 \implies (j - t)(j + t) - (j - t) = (j - t)(j + t - 1) = 7450.$

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.

$t^4 - 5200t^2 + t + 6,749,950 = 0.$ I have no idea why this does not render properly.

Factor the constant term with the absolutely obvious factor of 10

$6,749,950 = 10 *674,995.$

Factor again by the obvious factor of 5

$6,749,950 = 10 * 5 * 134999.$

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.

$50^4 - 5200 * 50^2 + 50 = 6,250,000 - 50 * 2500 + 50 =$

$6,250,050 - 13,000,000 = -\ 6,749,950.$

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.
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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