Show that the following equation has no rational solutions: (x^29 +1)^4 ?(x^29 +

Question

Show that the following equation has no rational solutions: (x^29 +1)^4 ?(x^29 +1)+1=0

Explanation / Answer

For a quadratic equation ax^2 + bx + c = 0, you can find the roots using the formula x = { [-b + sqrt( b^2 - 4ac )] / 2a} or x = { [-b - sqrt( b^2 - 4ac )] / 2a}..........[A quadratic equation has two roots] If b^2 - 4ac < 0 , then the roots are imaginary. (Since the square root of a negative number is imaginary) If b^2 - 4ac = 0 , then the roots are real and equal. (As sqrt of b^2 - 4ac = 0, the roots will simply be given by -b/2a ) If b^2 - 4ac > 0 and also a perfect square (like 4, 25, 1, 64, etc) , then the roots are real and rational. If b^2 - 4ac > 0 and not a perfect square (like 2, 7, 24, 50, etc) , then the roots are real and irrational. Now in your second exercise the nature of the roots is already given, so all you have to do is equate b^2 - 4ac with the correct condition from the ones given above. These are the basics that will get you started on these problems. For example, 2x^2 -8x +8 =0 Here, a = 2, b = -8 and c = 8 b^2 - 4ac = 64 - (4)(2)(8) = 64 - 64 = 0 So, your roots are real and equal. You can check by substituting in the basic formula for roots: x = { [-b + sqrt( b^2 - 4ac )] / 2a} or x = { [-b - sqrt( b^2 - 4ac )] / 2a} ie, we get, x = 2 or x = 2. ie, both the roots are same or equal and are real numbers. Another example: 2x^2 +6x +k = 0 find the smallest integral value of k, such that the equation has imaginary roots. Here, a = 2, b = 6 and c = k b^2 - 4ac < 0 (since the roots are imaginary). therefore, b^2 - 4ac = 36 - (4*2*k) < 0 ie, 36 - 8k < 0 ie, 36 < 8k or 8k > 36 therefore, k>4 ie, For all values of k that are greater than 4, the roots of the quadratic will be imaginary numbers. So, the smallest integral value of k such that the equation has imaginary roots is 5. (You can verify this by substituting k = 4 and k = 5 in the equation and finding the roots using the formula)

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