There's a method that doesn't require you to find the roots themselves. Call the unknown roots `r_1,r_2.` We know that the quadratic factors as

`x^2-6x+8=(x-r_1)(x-r_2),` and expanding the right side gives

`x^2-6x+8=x^2-(r_1+r_2)x+r_1r_2.` Equating coefficients gives

`r_1+r_2=6` **(1)**

`r_1r_2=8` **(2)**

Multiply equation **(1)** by `r_1` and then `r_2` to get the new equations

`r_1^2+r_1r_2=6r_1` **(3)**

`r_1r_2+r_2^2=6r_2` **(4)**

Add **(3)** and **(4)** to get `r_1^2+r_2^2+2r_1r_2=6(r_1+r_2).`

Now use **(1)** and **(2)** to get `r_1^2+r_2^2+16=36,` **so we get the answer `r_1^2+r_2^2=20.` **

*For this particular problem*, this method takes longer. However, it can be used to solve similar problems where solving for the roots is tedious.

The quadratic equation given is x^2 - 6x + 8 = 0.

x^2 - 6x + 8 = 0

=> x^2 - 4x - 2x + 8 = 0

=> x(x - 4) - 2(x - 4) = 0

=> (x - 2)(x - 4) = 0

=> x = 2 and x = 4

The sum of the squares of the roots is 2^2 + 4^2 = 4 + 16 = 20

**The required sum is 20.**

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