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Bunuel pls explain this question.
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Rikhraj1
Bunuel pls explain this question.
Since the roots are reciprocal of each other, the product of 2 roots (\(\frac{1}{r1} * r1\)) will always be 1.

Now whenever a quadratic equation of the form \(ax^2 + bx + c = 0\), where a, b and c are constants is given, the product of its roots is given by \(\frac{c}{a}\)

In the equation given to us, our \(\frac{c}{a}\) is \(\frac{2}{p}\), and since we already know that product of roots is 1 => 2/p = 1 => p = 2

Here's theory on sum and product of roots
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Krunaal
Rikhraj1
Bunuel pls explain this question.
Since the roots are reciprocal of each other, the product of 2 roots (\(\frac{1}{r1} * r1\)) will always be 1.

Now whenever a quadratic equation of the form \(ax^2 + bx + c = 0\), where a, b and c are constants is given, the product of its roots is given by \(\frac{c}{a}\)

In the equation given to us, our \(\frac{c}{a}\) is \(\frac{2}{p}\), and since we already know that product of roots is 1 => 2/p = 1 => p = 2

Here's theory on sum and product of roots
Krunaal sorry, still unable to understand it.
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Krunaal
Rikhraj1
Bunuel pls explain this question.
Since the roots are reciprocal of each other, the product of 2 roots (\(\frac{1}{r1} * r1\)) will always be 1.

Now whenever a quadratic equation of the form \(ax^2 + bx + c = 0\), where a, b and c are constants is given, the product of its roots is given by \(\frac{c}{a}\)

In the equation given to us, our \(\frac{c}{a}\) is \(\frac{2}{p}\), and since we already know that product of roots is 1 => 2/p = 1 => p = 2

Here's theory on sum and product of roots
Krunaal sorry, still unable to understand it.
Can you please share which part you're not able to follow? I can then zoom into it and try to explain.
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Rikhraj1
Krunaal sorry, still unable to understand it.
Can you please share which part you're not able to follow? I can then zoom into it and try to explain.
i m not able to relate "the product of 2 roots (1/r1∗r1) will always be 1" with "the product of its roots is given by c/a".
Therefore, didn't understand this whole part "in the equation given to us, our c/a is 2/p 2, and since we already know that product of roots is 1 => 2/p = 1 => p = 2".
And this my 1st time to encounter this concept.
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Krunaal
Rikhraj1
Krunaal sorry, still unable to understand it.
Can you please share which part you're not able to follow? I can then zoom into it and try to explain.
i m not able to relate "the product of 2 roots (1/r1∗r1) will always be 1" with "the product of its roots is given by c/a".
Therefore, didn't understand this whole part "in the equation given to us, our c/a is 2/p 2, and since we already know that product of roots is 1 => 2/p = 1 => p = 2".
And this my 1st time to encounter this concept.

Comparing general form of quadratic equation: \(ax^2 + bx + c = 0\) with \(px^2 + qx + 2 = 0\); we get a = p; b = q; c = 2 .......... (i)

To calculate product of roots of \(ax^2 + bx + c = 0\), we have a formula => \(\frac{c}{a}\)......(ii) (You can memorize this formula, but if you're curious to know how it is derived, you can refer here)

Now from (i) and (ii), you can say product of roots = \(\frac{2}{p}\) .......(iii)

The question also says that the roots are reciprocal of each other, thus when you multiply the roots you will always get 1 => product of roots = 1 .......(iv)

Now from (iii) and (iv), 2/p = 1
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