Question

find the zeros of the polynomial 6√2x^2+13x+3√2

Answer 1

$6 \sqrt{ 2} {x}^{2} + 13x + 3 \sqrt{2} = 0 \\ 6 \sqrt{2} {x}^{2} + 9x + 4x + 3 \sqrt{2} = 0 \\ 3x(2 \sqrt{2}x + 3) + 2 \times\sqrt{2} \times \sqrt{2} x \: +3 \sqrt{2} = 0 \\ 3x(2 \sqrt{2} x + 3) + \sqrt{2} (2 \sqrt{2}x + 3) = 0 \\ (2 \sqrt{2}x + 3)(3x + \sqrt{2} ) = 0 \\ 2 \sqrt{2} x + 3 = 0 \\ x = \frac{ - 3}{2 \sqrt{2} } \: \: \: \: ans \\ you \: can \: rationalise \: the \: denominator \\ x = \frac{ - 3 \sqrt{2} }{4} \: \: \: \: ans \\ 3x + \sqrt{2} = 0 \\ x = \frac{ - \sqrt{2} }{3} \: \: \: \: \: ans$