Question

6x^2+ 12x-18=0 sum product and roots

Answer 1

at the end of 1 year on Rupees 1 and 1808 8% per annum compounded annually

Answer 2

$hey \: mate \: here \: is \: your \: answer...$
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$6 {x}^{2} + 12x - 18 = 0 \\ 6 {x}^{2} + (18 - 6)x - 18 = 0 \\ {6x}^{2} + 18x - 6x - 18 = 0 \\ 6x(x + 3) - 6(x + 3) = 0 \\ (6x - 6)(x + 3) = 0 \\ 6x - 6 = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x + 3 = 0 \\ \: \: \: \: \: \: \: 6x = 6 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x = - 3 \\ \: \: \: \: \: \: \: \: \: x = \frac{6}{6} \\ \: \: \: \: \: \: \: \: \: x = 1$

$sum \: of \: zeros \: = \frac{ - b}{a} \\ \\ 1 + ( - 3) = \frac{ - 12}{6} \\ \\ \: \: \: \: \: \: \: 1 - 3 = - 2 \\ \: \: \: \: \: \: \: \: \: \: \: - 2 = - 2$

$product \: of \: zeros = \frac{c}{a} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 1 \times - 3 = \frac{ - 18}{6} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: - 2 = - 2$

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$hope \: it \: helps.....$