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27 X 2 − 10 + 1 = 0 - Mathematics

\[27 x^2 - 10 + 1 = 0\]

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Solution

Given: 

\[27 x^2 - 10x + 1 = 0\]

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

\[a x^2 + bx + c = 0\], we get
\[a = 27, b = - 10\] and \[c = 1\] .
Substituting these values in
\[\alpha = \frac{- b + \sqrt{b^2 - 4ac}}{2a}\] and \[\beta = \frac{- b - \sqrt{b^2 - 4ac}}{2a}\],we get:
\[\alpha = \frac{10 + \sqrt{100 - 4 \times 27 \times 1}}{2 \times 27}\] and \[\beta = \frac{10 - \sqrt{100 - 4 \times 27 \times 1}}{2 \times 27}\]
\[\Rightarrow \alpha = \frac{10 + \sqrt{100 - 108}}{54}\]   and   \[\beta = \frac{10 - \sqrt{100 - 108}}{54}\]
\[\Rightarrow \alpha = \frac{10 + \sqrt{- 8}}{54}\]  and  \[\beta = \frac{10 - \sqrt{- 8}}{54}\]
\[\Rightarrow \alpha = \frac{10 + \sqrt{8 i^2}}{54}\] and \[\beta = \frac{10 - \sqrt{8 i^2}}{54}\]
\[\Rightarrow \alpha = \frac{10 + i2\sqrt{2}}{54}\] and   \[\beta = \frac{10 - i2\sqrt{2}}{54}\]
\[\Rightarrow \alpha = \frac{2(5 + i\sqrt{2})}{54}\]    and \[\beta = \frac{2(5 - i\sqrt{2})}{54}\]
\[\Rightarrow \alpha = \frac{5}{27} + \frac{\sqrt{2}}{27}i\] and \[\beta = \frac{5}{27} - \frac{\sqrt{2}}{27}i\]
Hence, the roots of the equation are 
\[\frac{5}{27} \pm \frac{\sqrt{2}}{27}i\] .
  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 14 | Page 6
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