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प्रश्न
Solve the following quadratic equation:
\[2 x^2 + \sqrt{15}ix - i = 0\]
उत्तर
\[ 2 x^2 + \sqrt{15} ix - i = 0\]
\[\text { Comparing the given equation with the general form } a x^2 + bx + c = 0, \text { we get }\]
\[a = 2, b = \sqrt{15} i \text { and } c = - i\]
\[x = \frac{- b \pm \sqrt{b^2 - 4a c}}{2a}\]
\[ \Rightarrow x = \frac{- \sqrt{15} i \pm \sqrt{\left( \sqrt{15} i \right)^2 + 8i}}{4}\]
\[ \Rightarrow x = \frac{- \sqrt{15} i \pm \sqrt{8i - 15}}{4} . . . \left( i \right)\]
\[\text { Let }x + iy = \sqrt{8i - 15} . \text { Then }, \]
\[ \Rightarrow \left( x + iy \right)^2 = 8i - 15\]
\[ \Rightarrow x^2 - y^2 + 2ixy = 8i - 15 \]
\[ \Rightarrow x^2 - y^2 = - 15 \text { and } 2xy = 8 . . . \left( ii \right)\]
\[\text { Now, } \left( x^2 + y^2 \right)^2 = \left( x^2 - y^2 \right)^2 + 4 x^2 y^2 \]
\[ \Rightarrow \left( x^2 + y^2 \right)^2 = 225 + 64 = 289\]
\[ \Rightarrow x^2 + y^2 = 17 . . . \left( iii \right) \]
\[\text { From } \left( ii \right) \text { and } \left( iii \right)\]
\[ \Rightarrow x = \pm 1\text { and } y = \pm 4\]
\[\text { As, xy is positive } \left[ \text { From } \left( ii \right) \right]\]
\[ \Rightarrow x = 1, y = 4 \text { or, } x = - 1, y = - 4\]
\[ \Rightarrow x + iy = 1 + 4i \text { or,} - 1 - 4i\]
\[ \Rightarrow \sqrt{8i - 15} = \pm \left( 1 - 4i \right)\]
\[\text { Substituting these values in } \left( i \right), \text { we get }, \]
\[ \Rightarrow x = \frac{- \sqrt{15} i \pm \left( 1 + 4i \right)}{4} \]
\[ \Rightarrow x = \frac{1 + \left( 4 - \sqrt{15} \right)i}{4} , \frac{- 1 - \left( 4 + \sqrt{15} \right)i}{4}\]
\[\text { So, the roots of the given quadratic equation are } \frac{1 + \left( 4 - \sqrt{15} \right)i}{4} \text { and } \frac{- 1 - \left( 4 + \sqrt{15} \right)i}{4} . \]
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