Question

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

Answer 1

`\frac a{b^2} + \frac b{a^2} = 10`

Explanation:

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

Answer 2

`(a)/(b^2)+(b)/(a^2)=10`

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=(-b \pm √(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 & =(-(-4) \pm √((-4)^2-4(1)(2)))/(2(1))\\& =(4 \pm √(8))/(2)\\& =(4 \pm 2√(2))/(2)\\& =2 \pm √(2)\end{aligned}`

`\textsf{Let }a=2+√(2)`

`\textsf{Let }b=2-√(2)`

Therefore:

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