Question

43) Find the zeroes of the polynomial
p(x) =√2x square-3x-2√2​

Answer 1

Step-by-step explanation:

\sqrt{2x {}^{2} } - 3x - 2 \sqrt{2} \\ \sqrt{2x {}^{2} } - (4x - x) + 2 \sqrt{2} \\ \sqrt{2x {}^{2} } - 4x + x - 2 \sqrt{2} \\ \sqrt{2x} (x - 2 \sqrt{2} + 1(x - 2 \sqrt{2} ) \\ ( \sqrt{2x} + 1)(x - 2 \sqrt{2} ) \\ \sqrt{2x} + 1 = 0 \\ x = - 1 \div \sqrt{2 } \\ x - 2 \sqrt{2 } = 0 \\ x = 2 \sqrt{2}

Answer 2

Step-by-step explanation:

$Step-by-step explanation: \\ </p><p>\sqrt{2x {}^{2} } - 3x - 2 \sqrt{2} \\ \sqrt{2x {}^{2} } - (4x - x) + 2 \sqrt{2} \\ \sqrt{2x {}^{2} } - 4x + x - 2 \sqrt{2} \\ \sqrt{2x} (x - 2 \sqrt{2} + 1(x - 2 \sqrt{2} ) \\ ( \sqrt{2x} + 1)(x - 2 \sqrt{2} ) \\ \sqrt{2x} + 1 = 0 \\ x = - 1 \div \sqrt{2 } \\ x - 2 \sqrt{2 } = 0 \\ x = 2 \sqrt{2}$