If (3+2 sqrt{2})^{x^2-4}+(3-2 sqrt{2})^{x^2-4 | vedclass

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(D) Given $(3+2 \sqrt{2}) \cdot (3-2 \sqrt{2}) = 9-8 = 1$. Thus,$(3-2 \sqrt{2}) = \frac{1}{3+2 \sqrt{2}}$. Let $y = (3+2 \sqrt{2})^{x^2-4}$. The equat

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$35$

Vedclass · Answer

(D) Given $(3+2 \sqrt{2}) \cdot (3-2 \sqrt{2}) = 9-8 = 1$. Thus,$(3-2 \sqrt{2}) = \frac{1}{3+2 \sqrt{2}}$. Let $y = (3+2 \sqrt{2})^{x^2-4}$. The equation becomes $y + \frac{1}{y} = 6$,which simplifies to $y^2 - 6y + 1 = 0$. Solving for $y$ using the quadratic formula,$y = \frac{6 \pm \sqrt{36-4}}{2} = 3 \pm 2 \sqrt{2}$. Case $1$: $(3+2 \sqrt{2})^{x^2-4} = 3+2 \sqrt{2}$ $\Rightarrow x^2-4 = 1$ $\Rightarrow x^2 = 5$. Case $2$: $(3+2 \sqrt{2})^{x^2-4} = 3-2 \sqrt{2} = (3+2 \sqrt{2})^{-1}$ $\Rightarrow x^2-4 = -1$ $\Rightarrow x^2 = 3$. If $x^2 = 5$,then $x^4+x^2+5 = (5)^2 + 5 + 5 = 25+5+5 = 35$. If $x^2 = 3$,then $x^4+x^2+5 = (3)^2 + 3 + 5 = 9+3+5 = 17$. Since $35$ is the only option provided,the correct answer is $35$.

If $(3+2 \sqrt{2})^{x^2-4}+(3-2 \sqrt{2})^{x^2-4}=6$,then $x^4+x^2+5=$

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