Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
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8 X 2 − 9 X + 3 = 0 - Mathematics

\[8 x^2 - 9x + 3 = 0\]

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Solution

Given: 

\[8 x^2 - 9x + 3 = 0\]

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

\[a x^2 + bx + c = 0\],  we get 
\[a = 8, b = - 9\] and \[c = 3\].
Substituting these values in 
\[\alpha = \frac{- b + \sqrt{b^2 - 4ac}}{2a}\] and \[\beta = \frac{- b - \sqrt{b^2 - 4ac}}{2a}\] we get:
\[\alpha = \frac{9 + \sqrt{81 - 4 \times 8 \times 3}}{2 \times 8}\]  and \[\beta = \frac{9 - \sqrt{81 - 4 \times 8 \times 3}}{2 \times 8}\]
\[\Rightarrow \alpha = \frac{9 + \sqrt{81 - 96}}{16}\] and  \[\beta = \frac{9 - \sqrt{81 - 96}}{16}\]
\[\Rightarrow \alpha = \frac{9 + \sqrt{- 15}}{16}\] and  \[\beta = \frac{9 - \sqrt{- 15}}{16}\]
\[\Rightarrow \alpha = \frac{9 + \sqrt{15 i^2}}{16}\] and \[\beta = \frac{9 - \sqrt{15 i^2}}{16}\]
\[\Rightarrow \alpha = \frac{9 + i\sqrt{15}}{16}\]  and \[\beta = \frac{9 - i\sqrt{15}}{16}\]
\[\Rightarrow \alpha = \frac{9}{16} - \frac{\sqrt{15}}{16}i\] and   \[\beta = \frac{9}{16} + \frac{\sqrt{15}}{16}i\]
Hence, the roots of the equation are \[\frac{9}{16} \pm \frac{\sqrt{15}}{16}i\] .
  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 17 | Page 6
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