If the equation x^8 - kx^2 + 3 = 0 has a real | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given the equation $x^8 - kx^2 + 3 = 0$,we can rewrite it as $k = \frac{x^8 + 3}{x^2} = x^6 + \frac{3}{x^2}$.To find the minimum value of $k$,we u

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) Given the equation $x^8 - kx^2 + 3 = 0$,we can rewrite it as $k = \frac{x^8 + 3}{x^2} = x^6 + \frac{3}{x^2}$.To find the minimum value of $k$,we use the Arithmetic Mean-Geometric Mean $(AM \geq GM)$ inequality.We split the term $\frac{3}{x^2}$ into three equal parts: $\frac{1}{x^2} + \frac{1}{x^2} + \frac{1}{x^2}$.Thus,$k = x^6 + \frac{1}{x^2} + \frac{1}{x^2} + \frac{1}{x^2}$.Applying $AM \geq GM$ for these four positive terms:$\frac{x^6 + \frac{1}{x^2} + \frac{1}{x^2} + \frac{1}{x^2}}{4} \geq \sqrt[4]{x^6 \cdot \frac{1}{x^2} \cdot \frac{1}{x^2} \cdot \frac{1}{x^2}} = \sqrt[4]{1} = 1$.Therefore,$x^6 + \frac{3}{x^2} \geq 4$.The minimum value of $k$ is $4$.

If the equation $x^8 - kx^2 + 3 = 0$ has a real solution,then the least integral value of $k$ is-

Similar Questions

If the numbers $a, b, c, d, e$ form an $A.P.$,then the value of $a - 4b + 6c - 4d + e$ is

Let $E = x^{2017} + y^{2017} + z^{2017} - 2017xyz$ (where $x, y, z \geq 0$),then the least value of $E$ is

The sum of the first $n$ terms of the series $\cot^{-1} 3 + \cot^{-1} 7 + \cot^{-1} 13 + \cot^{-1} 21 + \dots$ is given by:

If all roots of the equation $x^3 - 2ax^2 + 3bx - 8 = 0$ are positive,where $a, b \in R$,then the minimum value of $b$ is:

If $S_1, S_2, S_3, \dots, S_m$ are the sums of $n$ terms of $m$ arithmetic progressions $(A.P.)$ whose first terms are $1, 2, 3, \dots, m$ and common differences are $1, 3, 5, \dots, 2m - 1$ respectively,then $S_1 + S_2 + S_3 + \dots + S_m = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module