Correct option is (D) 4 – √3
Let the other root be \(\alpha.\)
Then sum of roots \(=\frac{-b}a\)
\(=\frac{-(-8)}1=8\)
\(\therefore \alpha+4+\sqrt3=8\)
\(\Rightarrow\alpha=8-(4+\sqrt3)\)
= \(4-\sqrt{3}\)
Hence, the other root of given quadratic equation is \(4-\sqrt{3}.\)
Correct option is (D) 2 – √3
Given quadratic equation is \(x^2-4x+1=0\)
Let the other root be x.
\(\therefore\) Sum of roots \(=\frac{-(-4)}1=4\)
\(\Rightarrow\) \(x+2+\sqrt{3}=4\)
\(\Rightarrow\) \(x=4-2-\sqrt3\)
\(\Rightarrow\) x = \(2-\sqrt{3}\)
Hence, the other root is \(2-\sqrt{3}.\)
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