Solving a system

[xxMsJojoxx]

How do I solve for x and y?


I got this, which is wrong.

 

[Dr.Peterson]

The next to last line is (or at least should be) [MATH]\frac{x^4+25}{x^2}=15[/MATH]. Your last line would only be correct if the next to last had been [MATH]\frac{x^4}{x^2}+25=15[/MATH], which is not equivalent. Do you see?

 

[Dr.Peterson]

Hi Ms Jojo. Your errors resulted from the incorrect way you deconstructed the ratio, in going from the second-to-last line to the last line. However, had you done that correctly, you still would have been stuck because of the fourth-degree polynomial result:

25 - 14x^4 = 0

I think that results from a typo. The fourth degree equation is [MATH]x^4-15x^2+25=0[/MATH], which can be solved by substitution (though the result is not pretty).

 

[Otis]

I've deleted my response, as I'd misread part of the image.

Instead, I'll provide an example for readers to consider.

\(\displaystyle \frac{1 + 2}{3} = 1\)

We're not allowed to subtract 2 from each side because 2 is joined together with 1 forming a numerator. Here's what happens, if we "subtract" the 2 on the left-hand side, by simply removing it from the ratio (as well as actually subtracting 2, from the right-hand side).

\(\displaystyle \frac{1}{3} = -1\)

Clearly, not true. If we really want to subtract, then we'd first need to deconstruct the given ratio into a sum of fractions having the common denominator 3.

\(\displaystyle \frac{1}{3} + \frac{2}{3} = 1\)

We may now subtract 2/3 from each side, obtaining a verity. I hope this example helps people who'd like to understand the subtraction error in the op.

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[xxMsJojoxx]

The next to last line is (or at least should be) [MATH]\frac{x^4+25}{x^2}=15[/MATH]. Your last line would only be correct if the next to last had been [MATH]\frac{x^4}{x^2}+25=15[/MATH], which is not equivalent. Do you see?

Thank you, Dr. Peterson. What is the best way, to solve for x? I'm kind of stuck, since none of the factors of 25 seem to make up the sum for -15.

 

[Otis]

… What is the best way, to solve for x? …

One way is to make a substitution, instead of trying to factor (as DrP hinted). Here's another hint:

The polynomial ax^4 + bx^2 + c is quadratic in form. We can see that, by picking a symbol to represent x^2, followed by making the substitution. For example,

Let x^2 = t

Substituting t for x^2, we obtain a quadratic polynomial, and we could then use the Quadratic Formula to find that new polynomial's roots (in terms of t).

at^2 + bt + c = 0

Once we have solutions for t, we "back substitute" them for t (one at a time) into our equation x^2=t, followed by solving for x, to find x-values that are roots of the original 4th-degree polynomial.

Polynomials like ax^6 + bx^3 + c and ax^8 + bx^4 + c are also quadratic in form; we can make appropriate substitutions.

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[Dr.Peterson]

As I said, it isn't pretty. This can't be factored; I had to use the quadratic formula to solve for x^2, and then take the square root of that (taking into account the fact that y<0, and therefore also x<0).

I would have expected a nicer solution in an assigned problem, but this happens in real life!

 

[Steven G]

I would note that (x+y)^2 = (x^2 + y^2) + 2(xy) = 25

So x+y = +/- 5. Now since y<0 along with xy=5 implies that x<0. If you add two negative numbers you will get a negative number. So x+y=-5

xy=5 so y = 5/x

x + 5/x = -5
Multiply both sides by x yields x^2 - 5x + 5 = 0. Solve this equation for x and recall that x<0. Then solve for y.

 

[Otis]

Hi, again. Readers who would like to see an example of such substitution (obtaining a quadratic polynomial from a given polynomial that's quadratic in form) may watch the first 1¼ minutes of this video.

The remainder of the example is solved by factoring, but could also be solved using the Quadratic Formula. I'll post an example of the back substitution, later.

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[Otis]

… x^2 - 5x + 5 = 0 …

Solve … for x and recall that x<0. Then solve for y.

Hey Jomo. By "recall that x<0", you're saying to change the sign on each of those solutions, before substituting for x in y=5/x, right?

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[Otis]

In the video example (post #9), the quadratic equation they end up with is

u^2 - 20u + 64 = 0

That polynomial factors nicely, so the rest of the video used a factoring approach. In cases where a quadratic polynomial doesn't factor nicely, we may use the Quadratic Formula or Complete the Square, instead.

Here's how the video would have finished, were they to have used the Quadratic Formula (instead of factoring):

The Quadratic Formula yields u = 4 or u = 16

Next, back-substitute each of those solutions into the equation they wrote when first making the u-substitution. Then solve for x.

u = x^2
4 = x^2
x = 2 or x = -2

u = x^2
16 = x^2
x = 4 or x = -4

We obtain the same four roots, for x^4 - 20x^2 + 64, as they did in the example.

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[Otis]

… I would have expected a nicer solution in an assigned problem, but this happens in real life!

At least it wasn't an exam problem, heh.

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For all readers, changing the first equation in the op to x^2+y^2=26 yields a nicer solution. Here's a worked solution, for that example. The actual exercise may be solved the same way.

As before, y = 5/x and we're still interested only in solutions where x and y are both negative (i.e., solutions in Quadrant III).

x^2 + (5/x)^2 = 26

x^2 + 25/x^2 = 26

x^4 + 25 = 26x^2

x^4 - 26*x^2 + 25 = 0

Let u = x^2 (then u^2 = x^4), and make the substitutions.

u^2 - 26u + 25 = 0

The left-hand side factors, but I'll use the Quadratic Formula to solve for u, since that's what we'd do in the actual exercise.

u = [26 ± √( (-26)^2 - 4(1)(25) )] / [2(1)]

The Discriminant ( b^2 - 4ac ) evaluates to 576, which is 24^2. (This is the hallmark of quadratic polynomials that factor nicely: their Discriminant is always a perfect square.)

u = [26 ± 24] / 2

u = 13 ± 12

u = 25 or u = 1

Back-substituting these results into u = x^2, we obtain equations in terms of x.

25 = x^2
\(\;\) 1 = x^2

Solving those give solutions:

x = ±5 or
x = ±1

We know that y and x are related by y=5/x. We now use the negative x-values, to find corresponding y-values.

When x = -5:
y = 5/(-5)
y = -1

When x = -1:
y = 5/(-1)
y = -5

The solutions in Quadrant III are (-5,-1) and (-1,-5).

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Here's some extra information. All solutions of the system appear as intersection points of the graphs.

x^2 + y^2 = 26 is the equation of a circle with radius √26, centered at the Origin.

y = 5/x is a Rational function; its graph is very similar to the graph of y=1/x.



When we consider all four solutions, we see symmetry reflected in the values:

(1, 5) \(\quad\) (-1, -5)
(5, 1) \(\quad\) (-5, -1)

This is due to the symmetry of the graphs' shapes. The system in the op reflects the same sort of symmetry.

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[Steven G]

Hey Jomo. By "recall that x<0", you're saying to change the sign on each of those solutions, before substituting for x in y=5/x, right?

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You will get two real solutions. I am saying to just keep the solutions which are negative.
However upon further inspection that does not work. Yuck!
Thanks for pointing out your concern.
I need to think about this one!