Given: nbsp;21x2 #160;+ #160;9x #160;+ #160;1 #160;= #160;0 Comparing the given equation with nbsp;the general form of the quadratic equation ...

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Roots of a Quadratic Equation

21 × 2+9 x+1=...

Question

$21 x^{2} + 9 x + 1 = 0$

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Solution

Given: $21 x^{2} + 9 x + 1 = 0$
Comparing the given equation with the general form of the quadratic equation $a x^{2} + b x + c = 0$ , we get $a = 21, b = 9$ and $c = 1$ .

Substituting these values in $α = \frac{- b + \sqrt{b^{2} - 4 a c}}{2 a}$ and $β = \frac{- b - \sqrt{b^{2} - 4 a c}}{2 a}$ , we get:

$α = \frac{- 9 + \sqrt{81 - 4 \times 21 \times 1}}{2 \times 21}$ and $β = \frac{- 9 - \sqrt{81 - 4 \times 21 \times 1}}{2 \times 21}$

$\Rightarrow α = \frac{- 9 + \sqrt{3} i}{42}$ and $β = \frac{- 9 - \sqrt{3} i}{42}$

$\Rightarrow α = - \frac{9}{42} + \frac{\sqrt{3} i}{42}$ and $β = - \frac{9}{42} - \frac{\sqrt{3} i}{42}$

$\Rightarrow α = - \frac{3}{14} + \frac{\sqrt{3} i}{42}$ and $β = - \frac{3}{14} - \frac{\sqrt{3} i}{42}$

Hence, the roots of the equation are $- \frac{3}{14} \pm \frac{i \sqrt{3}}{42} .$

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