Question

If a and B are the zeroes of the polynomial 3x²+2x-6,then find the value of (1) a-B (2) a²-B² (3) a³+B³.

Answer 1

Solution:-

given by equation:-

3x²+2x-6

by formula

$\frac{ - b( + - ) \sqrt{ {b}^{2} - 4ac} }{2a} \\ \frac{ - 2( + - ) \sqrt{4 - 4 \times 3 \times ( - 6)} }{2 \times 3} \\ \frac{ - 2( + - ) \sqrt{4 + 72} }{6} \\ \frac{ - 2( + - )2 \sqrt{19} }{6} \\ \frac{ - 1( + - ) \sqrt{19} }{3} \\ a = \frac{ - 1 + \sqrt{19} }{3} \\ b = \frac{ - 1 - \sqrt{19} }{3} \\ \\ 1 \\( a - b) = \frac{ - 1 + \sqrt{19} }{3} - \frac{ - 1 - \sqrt{19} }{3} \\ \frac{ - 1 + \sqrt{19} + 1 + \sqrt{19} }{3} \\ (a - b) = \frac{2 \sqrt{19} }{3}$
2.
${a}^{2} - {b}^{2} = { (\frac{ - 1 + \sqrt{19} }{3} )}^{2} - {( \frac{ - 1 - \sqrt{19} }{3} )}^{2} \\ \frac{1 + 19 - 2 \sqrt{19} - 1 - 19 - 2 \sqrt{19} }{9} \\ ( {a}^{2} - {b}^{2} ) = \frac{ - 4 \sqrt{19} }{9}$
3.
${a}^{3} + {b}^{3} = {( \frac{ - 1 + \sqrt{19} }{3} )}^{3} + {( \frac{ - 1 - \sqrt{19} }{3} )}^{3} \\ \frac{ - 1 + 19 \sqrt{19} - 3 \sqrt{19}( - 1 + \sqrt{19} ) - 1 - 19 \sqrt{19} + 3 \sqrt{19} ( - 1 - \sqrt{19} ) }{9} \\( {a}^{3} + {b}^{3} ) = \frac{ - 2}{9} ans\\ i \: hope \: its \: help \:$