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mathmath333

  • one year ago

What does the following statement means ?

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} &\normalsize \text{If}\ a=1,\ b,c\in I\ \text{and the roots are rational numbers.}\hspace{.33em}\\~\\ &\normalsize \text{then the root must be an integer} \hspace{.33em}\\~\\ \end{align}}\)

  2. ParthKohli
    • one year ago
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    Oh, this. Look at the quadratic formula.\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]If \(a=1\):\[x = \frac{-b \pm \sqrt{b^2 - 4c}}{2}\]Now it says that the roots are rational, so basically \(b^2 - 4c\) is a perfect square. Now it's given that \(b,c\) are integers. If \(b\) is even, then \(b = 2k\) so the root must be numerator must be an even number (why?). Otherwise, if \(b\) is odd then \(b = 2k + 1\) so the root is odd, which is then added to an odd number and made even again. This makes the numerator even, and so the root of the quadratic equation is an integer in both cases.

  3. mathmath333
    • one year ago
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    how does \(b=2k\pm1\) makes the other root integer ?

  4. ParthKohli
    • one year ago
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    Sorry, I hurried through the reply. Let me restate that. It says that the roots are rational, so \(b^2 - 4c\) is a perfect square. Look at the numerator, i.e., \(-b \pm \sqrt{b^2 - 4c}\). If \(b= 2k \) then \(b^2 - 4c\) is obviously even, so its square root, which is an integer, is also even. The whole numerator is also even in that case. You can divide that by 2. Similarly for odd \(b = 2k +1\), so then \(b^2 - 4c\) is odd here, so its square root is odd. But \(-b \) is odd too, so you are adding an odd number to an odd number, making it even. So you can divide that by 2.

  5. mathmath333
    • one year ago
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    ok thnks

  6. mathmath333
    • one year ago
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    i have one more statement

  7. ParthKohli
    • one year ago
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    OK, go ahead.

  8. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} &\normalsize \text{if a quadratic equation in }\ x\ \text{has more than two roots} \hspace{.33em}\\~\\ &\normalsize \text{then it is an identity in }\ x. \hspace{.33em}\\~\\ \end{align}}\)

  9. ParthKohli
    • one year ago
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    Yeah, of course! The Fundamental Theorem of Algebra says that the number of complex roots of a polynomial equation is always equal to its degree. If you see a quadratic equation where three values of \(x\) satisfy the equation, then it will, in fact, work for any value of \(x\) (which is what an identity is).

  10. ParthKohli
    • one year ago
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    I remember this example:\[\dfrac{(x-a)(x-b)}{(c-a)(c-b)} + \frac{(x-b)(x-c)}{(a-b)(a-c)}+ \frac{(x-c)(x-a)}{(b-c)(b-a)}=1\]If you plug in \(a,b,c\), you can see that all of the three satisfy this equation. Without any further trouble, you can claim that this equation is an identity.

  11. mathmath333
    • one year ago
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    what does "an identity" means here

  12. ParthKohli
    • one year ago
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    An identity is something that is true for all values of the variable.\[\sin^2 x + \cos^2 x = 1\]The above is not an equation. It is an identity. It's true for all values of \(x\). You don't need to solve it.

  13. mathmath333
    • one year ago
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    are there any other examples of quadratic equation having more than 2 roots.

  14. ParthKohli
    • one year ago
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    Let me rewrite that... If a quadratic equation has more than two roots, then it has infinitely many roots.

  15. ParthKohli
    • one year ago
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    \[\frac{x^2 +x}{x} = x + 1\] is a "quadratic equation" with infinitely many roots.

  16. mathmath333
    • one year ago
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    ok thnx

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