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

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(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

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$4$

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(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-

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