Question

if a and b are zeroes of the quadratic polynomial 2x2 + 3x - 6, then find the values of (i) a2b plus ba2

Answer 1

Answer:

8¼

Step-by-step explanation:

$f(x) = 2x^2 + 3x - 6\\\\\text{Using quadratic formula (for derivation, see link in comments...)}\\\\\\x = \dfrac{-3 \pm \sqrt{9 + 48}}{2(2)}\\\\\\= \dfrac{-3 \pm \sqrt{57}}{4}\\\\\\\text{Thus two zeroes of f(x) are:}\\\\\\a = \dfrac{-3 + \sqrt{57}}{4}\\\\\\b = \dfrac{-3 - \sqrt{57}}{4}\\\\\\a^2b + b^2a\\\\\\= \left( \dfrac{-3 + \sqrt{57}}{4} \right) ^2 + \left( \dfrac{-3 - \sqrt{57}}{4} \right) ^2\\\\\\\\= \dfrac{(-3 + \sqrt{57})^2}{16} + \dfrac{(-3 - \sqrt{57})^2}{16}$

$= \dfrac{9 + 57 + 2(-3)(\sqrt{57}) + 9 + 57 - 2(-3)(\sqrt{57})}{16}\\\\\\\\= \dfrac{132}{16}\\\\\\\\= \large \boxed{\boxed{\boxed{\bold{8 \dfrac{1}{4}}}}}$

Answer 2

Answer:

Step-by-step explanation: