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

If $(3+2 \sqrt{2})^{x^2-4}+(3-2 \sqrt{2})^{x^2-4}=6$, then $x^4+x^2+5=$
  1. -30
  2. -35
  3. 30
  4. 35

Solution

Since $(3+2 \sqrt{2}) \cdot(3-2 \sqrt{2})=9-8=1$ $ \Rightarrow(3-2 \sqrt{2})=\frac{1}{(3+2 \sqrt{2})} $ Now $(3+2 \sqrt{2})^{x^2-4}+(3-2 \sqrt{2})^{x^2-4}=6$ $ \Rightarrow(3+2 \sqrt{2})^{x^2-4}+\frac{1}{(3+2 \sqrt{2})^{x^2-4}}=6 $ Let $y=(3+2 \sqrt{2})^{x^2-4}$ $ \Rightarrow y+\frac{1}{y}=6 \Rightarrow y^2-6 y+1=0 $ $\Rightarrow y=3+2 \sqrt{2}$ or $y=3-2 \sqrt{2}$ $\Rightarrow(3+2 \sqrt{2})^{x^2-4}=(3+2 \sqrt{2})$ or $ (3+2 \sqrt{2})^{x^2-4}=\frac{1}{(3+2 \sqrt{2})} $ $\Rightarrow x^2-4=1$ or $x^2-4=-1$ $\Rightarrow x=\sqrt{5}$ or $x=3 i$ Since $x=3 i$ is imaginary Hence $x=\sqrt{5}$ Therefore $x^4+x^2+5=35$

Asked in: AP EAMCET 2023 (15 May Shift 1)

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