Question

For what value(s) of 'a' quadratic equation 30 ax^2 - 6x + 1 = 0 has no real roots​

Answer 1

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Given:

• Quadratic equation : $p(x) = 30ax^2 - 6x + 1 = 0$

To Find:

• Value of 'a' such that p(x) has no real roots

Concept:

We know that, if a quadratic polynomial has no real roots, then the value of Discriminant(D) is less than 0.

Solution:

$\implies p(x) = 30ax^2 - 6x + 1 = 0 \\\\ D < 0 \\\\\implies B^2 - 4(A)(C) < 0 \\\\\text{Here, A = coefficient of x^2 , B = coefficient of x and C = constant.} \\\\\implies A = 30a, B = -6, C = 1 \\\\\implies (-6)^2 - 4(30a)(1) < 0 \\\\\implies 36 - 120a < 0 \\\\\implies 36 < 120a \\\\\implies 120a > 36 \\\\\implies a > \dfrac{36}{120} \\\\\implies \bf{ a > \dfrac{3}{10}}$

Conclusion:

Hence, for all values of $\bf{a > \dfrac{3}{10}}$, p(x) has no real roots.

Learn More:

• If, D > 0 => Real and distinct roots exist.
• If, D = 0 => Real and equal roots exist.
• If, D < 0 => No real roots exist. (imaginary roots exist)

Hope it helps u!!