Let p and q be two real numbers such that p + q = 3 and p4 + q4 = 369. Then
\((\frac{1}{p} + \frac{1}{q} )^{-2}\)
is equal to _______.
Let p and q be two real numbers such that p + q = 3 and p4 + q4 = 369. Then
\((\frac{1}{p} + \frac{1}{q} )^{-2}\)
is equal to _______.
Correct Answer: 4
Solution and Explanation
The correct answer is 4
∵ p + q = 3 …(i)
and p4 + q4 = 369 …(ii)
{(p + q)2 – 2pq}2 – 2p2q2 = 369
or (9 – 2pq)2 – 2(pq)2 = 369
or (pq)2 – 18pq – 144 = 0
∴ pq = –6 or 24
But pq = 24 is not possible
∴ pq = –6
Therefore
\((\frac{1}{p} + \frac{1}{q} )^{-2}\)
\(= ( \frac{pq}{p+q})\)
\( = ( -2)² = 4\)
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Concepts Used:
Quadratic Equations
A polynomial that has two roots or is of degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b, and c are the real numbers.
Consider the following equation ax²+bx+c=0, where a≠0 and a, b, and c are real coefficients.
The solution of a quadratic equation can be found using the formula, x=((-b±√(b²-4ac))/2a)
Two important points to keep in mind are:
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