Question 15351
 You posted so many questions without showing any work, 
 I don't quite feel happy to solve for you.

 1.	Write an equation in the form y=ax^2+bx+c for the quadratic function whose graph passes through (8,0),(0,8) and (-2,0).
Sol: Clearly, c = 8, then solve 
0 = 64a + 8b + 8 (or 8a + b = -1) and
0 = 4a – 2b (or b = 2a) 
     for a and b = ???

2.Find the roots of {{{ x^2 + (k^2+1)/k * x+1 }}}=0.
Sol: Use the quadratic formula directly.

3. If (2 /x – x/ 2)^2 = 0,evaluate x^6.
Sol  2 /x – x/ 2  = 0, so (4-x^2) /2x= 0, 
or x = (+/-) 2, Now x^6 = ???

4. Find all values of k that ensure that the roots are real for x-k(x-1)(x-2)=o.
Sol: Convert it to {{{-k x^2 + (3 k+1) x -2k }}} = 0 or
  {{{k x^2 - (3 k+1) x + 2k}}} = 0
Since if {{{ b^2 - 4ac = (3 k+1)^2 - 8k^2 >=0}}} , 
the given equation has only real roots
Implying {{{k^2 + 6k + 1 >= 0}}}, or {{{(k+3)^2 >=8}}}. 
Hence, {{{k >= -3+ 2 sqrt(2)}}} or 
 {{{ k <= -3 - 2 sqrt(2)}}}

5. Find all possible values of k so that 3x^2 + kx +5 can be factored as the product of two binomial factors with integer coefficients.
Sol: 3x^2 + kx +5 = (3x+1)(x+5) or (3x-1)(x-5) or (3x+5)(x+1) or (3x-5)(x-1).
Hence, k = ????

6. Show that there are nine pairs of positive integers (m,n) such that 
 {{{  m^2+3mn+2n^2-10m-20n=0.}}}
Sol: {{{m^2+3mn+2n^2-10m-20n}}} = (m + n)(m + 2n) –10(m+2n)
 = (m + 2n) (m+n-10). = 0.
Hence, m + 2n = 0 (no positive sol.)
Or m + n = 10, so there are 9 solutions 
{(1,9)(2,8),(3,7),(4,6),(5,5),(6,4),(3,7),(2,8)(1,9)}

7. The difference in the length of the hypotenuse of triangle ABC and the length of the hypotenuse of triangle XYZ is 3. Hypotenuse AB=x, hypotenuse XY= square root x-1 and AB>XY. Determine the length of each hypotenuse.
Sol: Solve the equation x – sqrt(x-1) = 3 or x-3 = sqrt(x-1) 
 So, (x-3)^2 = x –1 to get x^2-7x +10 = 0 and x = 2 or –5.(invalid)


 Bad notation:1. Never use "over" in fraction again. 
              2. Square root x-1 means ????

 You should work very hard.

 Kenny