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Solve the Following Quadratic Equation: 2 X 2 + √ 15 I X − I = 0 - Mathematics

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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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अध्याय 14: Quadratic Equations - Exercise 14.2 [पृष्ठ १३]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 14 Quadratic Equations
Exercise 14.2 | Q 2.07 | पृष्ठ १३

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