दिया है,
\(x^{2}-2 \sqrt{2} x+1=0\)
यहाँ
a = 1
\(b=-2 \sqrt{2}\)
\(c=1\)
\(D=b^{2}-4 a c\)
\(=(-2 \sqrt{2})^{2}-4 \times 1 \times 1\)
\(=8-4\)
\(=4\)
\( x=\frac{-b \pm \sqrt{D}}{2 a}\)
\(=\frac{-(-2 \sqrt{2}) \pm \sqrt{4}}{2 \times 1} ;\)
\(x=\frac{2 \sqrt{2} \pm 2}{2} \)
\(x=\frac{2(\sqrt{2}+1)}{2}\) या
\(x=\frac{2(\sqrt{2}-1)}{2} \)
\(\therefore x=\sqrt{2}+1\) या \(\sqrt{2}-1 \)
