if a and b are non-zero distinct real numbers, determine the nature of the roots of the following quadratic equations.

a)  x 2次+(b-a) x + (a-b)2次=0

b) -2(a+1) x 2次+(a-4) x +3a-2=0 for a 不等於 -1

a的answer 系 no real roots

b的answer系 two unequal real roots

话说我数学考过很多次不及格...

  还是算了...

我中4..你中4...你d數學咁可怕的

如果比中文我睇的話..我可能幫到你

唉 聽朝借黎抄

我連最易都唔識做

我第一條亂做算= =

第2條..研究緊..

a)  x 2次+(b-a) x + (a-b)2次=0

b) -2(a+1) x 2次+(a-4) x +3a-2=0 for a 不等於 -1

a .                 x^2+xb-xa+a^2-2ab+b^2=0?  唔太明你打緊乜.一係你用小畫家寫出黎

I help you for (a)

consider sum of roots: -b+a=b-a+(a+b)square

                       2(a-b)=(a-b)square

                         a-b=2

consider discriminent

(b-a)square - 4(1)(a-b)square

b square - 2ab + a square -4(a square - 2ab + b square)

-3a square - 3b square + 6ab

-3(a square -2ab + b square)

-3(a-b) square

subs a-b=2

D=-12

Since D<0

so this equation has no real roots

x就係x..............我都唔記得乜野係real roots, no real