0
$\begingroup$

Why is there no solution for the system of equations with a greater than 0?

Solve[{c^2 - b^4/a^2 == 0, a^2 + b^2 == c^2, e == c/a, a > 0, 
  e > 1}, e, {a, c}]

Why can we solve the value of e by setting a not equal to 0?

Solve[{c^2 - b^4/a^2 == 0, a^2 + b^2 == c^2, e == c/a, a != 0, 
  e > 1}, e, {a, c}]

{{e -> 1/2 (1 + Sqrt[5])}}

How can we solve this equation if a is greater than 0?

Solve[{(((2 Sqrt[2] + 2) a)^2 + 8 a^2 - 4 c^2)/(
   2 (2 Sqrt[2] + 2) a 2 Sqrt[2] a) == Sqrt[2]/2, e == c/a, a > 0, 
  e > 1}, e, {a, c}]

the result is:

{{e -> Sqrt[3]}}
Improve this question
$\endgroup$
3
  • 2
    $\begingroup$ Solve[{c^2 - b^4/a^2 == 0, a^2 + b^2 == c^2, e == c/a, a > 0, e > 1}, e, {a, c}, Reals] $\endgroup$
    – cvgmt
    Mar 12 at 2:54
  • 1
    $\begingroup$ The result of Solve[{c^2 - b^4/a^2 == 0, a^2 + b^2 == c^2, e == c/a, a > 0, e > 1}, e, {a, b}, Reals] is more informative. $\endgroup$
    – user64494
    Mar 12 at 6:59
  • 1
    $\begingroup$ Looks like a bug to me. I would report it to "support@wolfram.com" $\endgroup$
    – Daniel Huber
    Mar 12 at 9:02

2 Answers 2

Reset to default
1
$\begingroup$

There are no solutions with generic complex b -- Im[b] needs to be zero.

In[1]:= Solve[{c^2 - b^4/a^2 == 0, a^2 + b^2 == c^2, e == c/a, a > 0, e > 1}, e, {a,
 c}, MaxExtraConditions->1]//InputForm                                              

Out[1]//InputForm= 
{{e -> ConditionalExpression[(1 + Sqrt[5])/2, (Im[b] == 0 && Re[b] > 0) || 
     (Im[b] == 0 && Re[b] < 0)]}}

There is a solution with generic real b.

In[2]:= Solve[{c^2 - b^4/a^2 == 0, a^2 + b^2 == c^2, e == c/a, a > 0, e > 1}, e, {a,
 c}, Reals]//InputForm                                                              

Out[2]//InputForm= {{e -> (1 + Sqrt[5])/2}}
Improve this answer
$\endgroup$
3
  • $\begingroup$ Solve[{c^2 - b^4/a^2 == 0, a^2 + b^2 == c^2, e == c/a, a != 0, e > 1}, e, {a, b, c}] $\endgroup$
    – csn899
    Mar 13 at 5:57
  • $\begingroup$ Solve[{c^2 - b^4/a^2 == 0, a^2 + b^2 == c^2, e == c/a, a > 0, e > 1}, e, {a, b, c}] $\endgroup$
    – csn899
    Mar 13 at 5:57
  • $\begingroup$ After your explanation, we know that the solution of this equation is related to whether the parameter b is a complex number whose imaginary part is not equal to 0. The value of e can be calculated successfully in the above two forms. {a,b,c} $\endgroup$
    – csn899
    Mar 13 at 6:00
2
$\begingroup$

I opine that your original problem was poorly stated.

I took the liberty of restating yor problem after noting the $e$ could computed afterwards:

solution = Reduce[{c^2 - b^4/a^2 == 0, a^2 + b^2 == c^2, 
c/a > 0, a > 0}, {a, b, c}, Reals]

I used Reduce to attempt to avail all the solutions.

ConditionalExpression[{ToRules[solution[[2]]]}, a > 0]

$$\fbox{$\left\{\left\{b\to -\sqrt{\frac{1}{2} \sqrt{5} \sqrt{a^4}+\frac{a^2}{2}},c\to \frac{b^2}{a}\right\}, \\ \left\{b\to \sqrt{\frac{1}{2} \sqrt{5} \sqrt{a^4}+\frac{a^2}{2}},c\to \frac{b^2}{a}\right\}\right\}\text{ if }a>0$}$$

ConditionalExpression[{{b -> -Sqrt[a^2/2 + (1/2)*Sqrt[5]*Sqrt[a^4]], c -> b^2/a}, {b -> Sqrt[a^2/2 + (1/2)*Sqrt[5]*Sqrt[a^4]], c -> b^2/a}}, a > 0]

Of course, I may have missed something.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.