MCQ
If $\text{p}(\text{x})=\text{x}^2-2\sqrt{2}\text{x}+1,$ than $\text{p}(2\sqrt{2})$ is equal to:
- A$0$
- B$4\sqrt{2}$
- C$8\sqrt{2}+1$
- ✓$1$
✓
Answer
Correct option: D.
$1$
$\text{p}(\text{x})=\text{x}^2-2\sqrt{2}\text{x}+1,$
$\Rightarrow\text{p}(2\sqrt{2})=(2\sqrt{2})^2-2\sqrt{2}(2\sqrt{2})+1$
$\Rightarrow\text{p}(2\sqrt{2})=8-8+1$
$\Rightarrow\text{p}(2\sqrt{2})=1$
$\Rightarrow\text{p}(2\sqrt{2})=(2\sqrt{2})^2-2\sqrt{2}(2\sqrt{2})+1$
$\Rightarrow\text{p}(2\sqrt{2})=8-8+1$
$\Rightarrow\text{p}(2\sqrt{2})=1$
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