Question

Let a and b be roots of x² - 4x + 2 = 0. find the value of a/b² +b/a²​

Answer 1

Answer:

$\dfrac{a}{b^2}+\dfrac{b}{a^2}=10$

Step-by-step explanation:

Given equation:   $x^2-4x+2=0$

The roots of the given quadratic equation are the values of x when $y=0$.

To find the roots, use the quadratic formula:

$x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0$

Therefore:

$a=1, \quad b=-4, \quad c=2$

$\begin{aligned}\implies x & =\dfrac{-(-4) \pm \sqrt{(-4)^2-4(1)(2)}}{2(1)}\\& =\dfrac{4 \pm \sqrt{8}}{2}\\& =\dfrac{4 \pm 2\sqrt{2}}{2}\\& =2 \pm \sqrt{2}\end{aligned}$

$\textsf{Let }a=2+\sqrt{2}$

$\textsf{Let }b=2-\sqrt{2}$

Therefore:

$\begin{aligned}\implies \dfrac{a}{b^2}+\dfrac{b}{a^2} & = \dfrac{2+\sqrt{2}}{(2-\sqrt{2})^2}+\dfrac{2-\sqrt{2}}{(2+\sqrt{2})^2}\\\\& = \dfrac{2+\sqrt{2}}{6-4\sqrt{2}}+\dfrac{2-\sqrt{2}}{6+4\sqrt{2}}\\\\& = \dfrac{(2+\sqrt{2})(6+4\sqrt{2})+(2-\sqrt{2})(6-4\sqrt{2})}{(6-4\sqrt{2})(6+4\sqrt{2})}\\\\& = \dfrac{12+8\sqrt{2}+6\sqrt{2}+8+12-8\sqrt{2}-6\sqrt{2}+8}{36+24\sqrt{2}-24\sqrt{2}-32}\\\\& = \dfrac{40}{4}\\\\& = 10\end{aligned}$

Answer 2

Answer:

$\dfrac a{b^2} + \dfrac b{a^2} = 10$

Step-by-step explanation:

$\text{Given that, the roots are a,b and } ~ x^2 -4x+2 = 0\\\\\text{So,}\\\\a+b = -\dfrac{-4}1 = 4\\\\ab = \dfrac 21 = 2\\\\\text{Now,}\\\\~~~~~\dfrac a{b^2} + \dfrac b{a^2}\\\\\\=\dfrac{a^3 +b^3}{a^2b^2}\\\\\\=\dfrac{(a+b)^3 -3ab(a+b)}{(ab)^2}\\\\\\=\dfrac{4^3 -3(2)(4)}{2^2}\\\\\\=\dfrac{64-24}{4}\\\\\\=\dfrac{40}{4}\\\\\\=10$