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√ 5 X 2 + X + √ 5 = 0 - Mathematics

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प्रश्न

\[\sqrt{5} x^2 + x + \sqrt{5} = 0\]

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उत्तर

Given: 

\[\sqrt{5} x^2 + x + \sqrt{5} = 0\]

Comparing the given equation with the general form of the quadratic equation 

\[a x^2 + bx + c = 0\], we get
\[a = \sqrt{5} , b = 1\] and \[c = \sqrt{5}\].
Substituting these values in
\[\alpha = \frac{- b + \sqrt{b^2 - 4ac}}{2a}\] and \[\beta = \frac{- b - \sqrt{b^2 - 4ac}}{2a}\], we get:
\[\alpha = \frac{- 1 + \sqrt{1 - 4 \times \sqrt{5} \times \sqrt{5}}}{2\sqrt{5}}\]   and \[\beta = \frac{- 1 - \sqrt{1 - 4 \times \sqrt{5} \times \sqrt{5}}}{2\sqrt{5}}\]
\[\alpha = \frac{- 1 + \sqrt{- 19}}{2\sqrt{5}}\] and  \[\beta = \frac{- 1 - \sqrt{- 19}}{2\sqrt{5}}\]
\[\alpha = \frac{- 1 + i\sqrt{19}}{2\sqrt{5}}\] and \[\beta = \frac{- 1 - i\sqrt{19}}{2\sqrt{5}}\]
Hence, the roots of the equation are 
\[\frac{- 1 \pm i\sqrt{19}}{2\sqrt{5}}\].
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अध्याय 14: Quadratic Equations - Exercise 14.1 [पृष्ठ ६]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 14 Quadratic Equations
Exercise 14.1 | Q 24 | पृष्ठ ६
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