mathmath333
  • mathmath333
What does the following statement means ?
Mathematics
  • Stacey Warren - Expert brainly.com
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
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mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} &\normalsize \text{If}\ a=1,\ b,c\in I\ \text{and the roots are rational numbers.}\hspace{.33em}\\~\\ &\normalsize \text{then the root must be an integer} \hspace{.33em}\\~\\ \end{align}}\)
ParthKohli
  • ParthKohli
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.
mathmath333
  • mathmath333
how does \(b=2k\pm1\) makes the other root integer ?

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ParthKohli
  • ParthKohli
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.
mathmath333
  • mathmath333
ok thnks
mathmath333
  • mathmath333
i have one more statement
ParthKohli
  • ParthKohli
OK, go ahead.
mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} &\normalsize \text{if a quadratic equation in }\ x\ \text{has more than two roots} \hspace{.33em}\\~\\ &\normalsize \text{then it is an identity in }\ x. \hspace{.33em}\\~\\ \end{align}}\)
ParthKohli
  • ParthKohli
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).
ParthKohli
  • ParthKohli
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.
mathmath333
  • mathmath333
what does "an identity" means here
ParthKohli
  • ParthKohli
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.
mathmath333
  • mathmath333
are there any other examples of quadratic equation having more than 2 roots.
ParthKohli
  • ParthKohli
Let me rewrite that... If a quadratic equation has more than two roots, then it has infinitely many roots.
ParthKohli
  • ParthKohli
\[\frac{x^2 +x}{x} = x + 1\] is a "quadratic equation" with infinitely many roots.
mathmath333
  • mathmath333
ok thnx

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