Question

Find the zeros of √5x2+9x+4√5

Answer 1

We have,

$\quad \sqrt{5} {x}^{2} + 9x + 4 \sqrt{5} \\ \\ = \sqrt{5} {x}^{2} + (5 + 4)x + 4 \sqrt{5} \\ \\ = \sqrt{5} {x}^{2} + 5x + 4x + 4 \sqrt{5} \\ \\ = \sqrt{5} x(x) + \sqrt{5} x( \sqrt{5} ) + 4(x) + 4( \sqrt{5} ) \\ \\ = \sqrt{5} x(x + \sqrt{5} ) + 4(x + \sqrt{5} ) \\ \\ = (x + \sqrt{5} )( \sqrt{5} x + 4)$

For finding the zeroes of the polynomial,

$\texttt{let} \quad \sqrt{5} {x}^{2} + 9x + 4 \sqrt{5} = 0 \\ \\ \implies (x + \sqrt{5} )( \sqrt{5} x + 4) = 0 \\ \\ \implies x + \sqrt{5} = 0 \: \boxed{or} \: \sqrt{5} x + 4 = 0 \\ \\ \implies x = - \sqrt{5} \quad \: \boxed{or} \quad x = - \frac{4}{ \sqrt{5} }$