Question

let p and q are distinct naturals such that 2005+p=q^2 and 2005+q=p^2. find the value of 2014+pq

Answer 1

(2005+p)-(2005+q)=q^2-p^2
p-q=(q-p)(p+q)
p+q=(-1)
p^2+q^2=p+q+4010=4009
(p+q)^2=p^2+q^2+2pq=1
4009+2pq=1
pq=(-2004)
2014+pq=2014-2004=10

Answer 2

Solution :

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Given :

p² = 2005 + q

&

q² = 2005 + p

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To Find :

The value of :

⇒ 2014 + pq

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We know the identity that,

⇒ a² - b² = (a + b)(a - b)

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By subtracting q² from p²

We get,.

⇒ p² - q²

⇒ 2005 + q - (2005 + p)

⇒ 2005 + q - 2005 - p

⇒ q - p

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∴ p² - q² = q - p

⇒ (p + q)(p - q) = (q - p)

⇒ (p + q) (p - q) = - (p - q)

⇒ p + q = $\frac{-(p - q)}{p - q}$

⇒ p + q = -1 ......(i)

We know the identity that,

⇒ (a + b)² = a² + 2ab + b²

Hence,.

By substituting the values,

a = p & b = q,.

We get,

⇒ (p + q)² = p² + 2pq + q²

⇒ (-1)² = 2005 + q + 2pq + 2005 + p

⇒ 1 = 4010 + 2pq + p + q

⇒ 1 = 4010 + 2pq + (-1)

⇒ 1 = 4009 + 2pq

⇒ 1 - 4009 = 2pq

⇒ -4008 = 2pq

⇒ pq = $\frac{-4008}{2}$

⇒ pq = -2004

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Hence,.

⇒ 2014 + pq

⇒ 2014 + (-2004)

⇒ 2014 - 2004

⇒ 10

∴ 2014 + pq = 10

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                                          Hope it Helps !!