Given: 21x^{2} + 9x + 1 = 0

We shall apply discriminant rule,

Where, \(x= \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\)

Here, a = 21, b = 9, c = 1

So,

\( x=\dfrac{(-9 ±\sqrt{(9^2 – 4 (21)(1)))}}{2(21)}\)

= \( \dfrac{ (-9 ± \sqrt{3(-1))}}{42}\)

We have i^{2} = –1

By substituting –1 = i^{2} in the above equation, we get

\( x=\dfrac{(-9 ± \sqrt{3i^2)}}{42}\)

= \( -\dfrac{3}{14}±\sqrt{\dfrac{3i}{42}}\)

The roots of the given equation are \( -\dfrac{3}{14}±\sqrt{\dfrac{3i}{42}}\)

Answered by Aaryan | 1 year agoShow that 1 + i^{10} + i^{20} + i^{30} is a real number?

Solve the quadratic equations by factorization method only 6x^{2} – 17ix – 12 = 0

Solve the quadratic equations by factorization method only x^{2} + (1 – 2i)x – 2i = 0

Solve the quadratic equations by factorization method only x^{2} + 10ix – 21 = 0

Solve the quadratic equations by factorization method only 17x^{2} – 8x + 1 = 0