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For what value(s) of ‘a’ quadratic equation $30 a x^{2} - 6 x + 1 = 0$ has no real roots?

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a = \frac{3}{10}

a < \frac{3}{10}

a > \frac{1}{10}

a > \frac{3}{10}

answer is D.

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Detailed Solution

Given quadratic equation is

30 a x^{2} - 6 x + 1 = 0

.
For no real roots, D<0 where

D = b^{2} - 4 ac

.
Now comparing

30 a x^{2} - 6 x + 1 = 0

with

a x^{2} + bx + c = 0

we get,

a = 30 a, b = - 6, c = 1

Substituting the values in D,

\Rightarrow 0 > b^{2} - 4 ac

\Rightarrow 0 > {(- 6)}^{2} - 4 (30 a) (1)

\Rightarrow 0 > 36 - 120 a

\Rightarrow 120 a > 36

\Rightarrow a > \frac{36}{120}

\Rightarrow a > \frac{3}{10}

Thus, for all values of a greater than

\frac{3}{10},

the given quadratic equation will have imaginary roots.
Therefore, option (4) is correct.

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