if alpha and beta are the zeros of the polynomial x^2-3x-2 find alpha^3 beta^4+alfa^4 beta^3-alpha beta
Answers
Answer:
As it's given that \alphaα and \betaβ are the roots of the quadratic equation
x^2 -3x -2x2−3x−2
♦ By using Quadratic formula for finding roots .
= \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}=2a−b±b2−4ac
= \dfrac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-2)}}{2(1)}=2(1)−(−3)±(−3)2−4(1)(−2)
= \dfrac{3 \pm \sqrt{9 + 8}}{2}=23±9+8
= \dfrac{3 \pm \sqrt{17}}{2}=23±17
♦ So let
\alpha = \dfrac{3 + \sqrt{17}}{2}α=23+17
and
\beta = \dfrac{3 - \sqrt{17}}{2}β=23−17
♦ Now we will first find out the value of
>> (i)
\begin{lgathered}\dfrac{1}{2\alpha+\beta} \\\\ = \dfrac{1}{2\dfrac{3+\sqrt{17}}{2} + \dfrac{3-\sqrt{17}}{2}}\\\\=\dfrac{1}{\dfrac{6+2\sqrt{17}}{2} + \dfrac{3-\sqrt{17}}{2}}\\\\ = \dfrac{1}{9+\sqrt{17}}{2} \\\\ = \dfrac{2}{9+\sqrt{17}}\end{lgathered}2α+β1=223+17+23−171=26+217+23−171=9+1712=9+172
• On rationalizing denominator
\dfrac{2}{9+\sqrt{17}} \times \dfrac{9-\sqrt{17}}{9-\sqrt{17}}9+172×9−179−17
= \dfrac{18 - 2\sqrt{17}}{9^2 - \sqrt{17}^2 }=92−17218−217
\begin{lgathered}= \dfrac{18-2\sqrt{17}}{81-17} \\\\ = \dfrac{18-2\sqrt{17}}{64} \\\\ = \dfrac{9-\sqrt{17}}{32}\end{lgathered}=81−1718−217=6418−217=329−17
>> (ii)
= \dfrac{1}{2\beta + \alpha}=2β+α1
= \dfrac{1}{2\dfrac{3-\sqrt{17}}{2} + \dfrac{3+\sqrt{17}}{2}}=223−17+23+171
Step-by-step explanation: