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 _______.
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
\((\frac{1}{p} + \frac{1}{q} )^{-2}\)
\(= ( \frac{pq}{p+q})\)
\( = ( -2)² = 4\)
Let the lines $L_1 : \vec r = \hat i + 2\hat j + 3\hat k + \lambda(2\hat i + 3\hat j + 4\hat k)$, $\lambda \in \mathbb{R}$ and $L_2 : \vec r = (4\hat i + \hat j) + \mu(5\hat i + + 2\hat j + \hat k)$, $\mu \in \mathbb{R}$ intersect at the point $R$. Let $P$ and $Q$ be the points lying on lines $L_1$ and $L_2$, respectively, such that $|PR|=\sqrt{29}$ and $|PQ|=\sqrt{\frac{47}{3}}$. If the point $P$ lies in the first octant, then $27(QR)^2$ is equal to}
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)
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