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17 X 2 + 28 X + 12 = 0 - Mathematics

\[17 x^2 + 28x + 12 = 0\]

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Solution

Given: 

\[17 x^2 + 28x + 12 = 0\]

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

\[a x^2 + bx + c = 0\], we get
\[a = 17, b = 28\] and \[c = 12\]
Substituting these values in \[\alpha = \frac{- b + \sqrt{b^2 - 4ac}}{2a}\] and \[\beta = \frac{- b - \sqrt{b^2 - 4ac}}{2a}\] , we get: 
\[\alpha = \frac{- 28 + \sqrt{784 - 4 \times 17 \times 12}}{34}\]  and   \[\beta = \frac{- 28 - \sqrt{784 - 4 \times 17 \times 12}}{34}\]
\[\Rightarrow \alpha = \frac{- 28 + \sqrt{784 - 816}}{34}\] and  \[\beta = \frac{- 28 - \sqrt{784 - 816}}{34}\]
\[\Rightarrow \alpha = \frac{- 28 + \sqrt{- 32}}{34}\] and \[\beta = \frac{- 28 - \sqrt{- 32}}{34}\]
\[\Rightarrow \alpha = \frac{- 28 + \sqrt{32 i^2}}{34}\]    and \[\beta = \frac{- 28 - \sqrt{32 i^2}}{34}\]
\[\Rightarrow \alpha = \frac{- 28 + 4\sqrt{2} i}{34}\]  and \[\beta = \frac{- 28 - 4\sqrt{2} i}{34}\]
\[\Rightarrow \alpha = \frac{- 14 + 2\sqrt{2} i}{17}\]   and   \[\beta = \frac{- 14 - 2\sqrt{2} i}{17}\]
Hence, the roots of the equation are \[- \frac{14}{17} \pm \frac{2\sqrt{2}}{17}i .\]
  Is there an error in this question or solution?
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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 14 Quadratic Equations
Exercise 14.1 | Q 15 | Page 6
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