A MATH

1
x^2-k(x-1)=0
x^2-kx+k=0
discriminant=0
k^2-4k=0
k=0 or k=4
2
x^2-kx+k+3=0
discriminant=0
k^2-4(k+3)=0
k^2-4k-12=0
(k-6)(k+2)=0
k=6 or k=-2
3
px^2+2qx-p+2pq=0
discriminant
=4q^2-4p(-p+2q)
=4(p^2-2pq+q^2)
=4(p-q)^2
>=0
二次方程px^2+2qx-p+2p=0的根是有理數。
4
discriminant
=4(m+n)^2-8(m^2+n^2)
=4m^2+8mn+4n^2-8m^2-8n^2
=-4(m^2-2mn+n^2)
=-4(m-n)^2
<0 (since m not equal to n)
方程(m^2+n^2)x^2-2(m+n)x+2=0有非實數根
5
(x-a)(x-b)=c
x^2-(a+b)x+ab-c=0
discriminant
=(a+b)^2-4(ab-c)
=a^2+2ab+b^2-4ab+4c
=(a-b)^2+4c
>=0 (since c>0)
方程(x-a)(x-b)=c的根是實數

1. x^2 - kx + k = 0 , discriminant = 0
(-k)^2 - 4(k) = 0
k(k - 4) = 0
k = 0 or 4

2. x^2 - kx + k + 3 = 0 , discriminant = 0
(-k)^2 - 4(k+3) = 0
k^2 - 4k - 12 = 0
(k + 2)(k - 6) = 0
k = -2 or 6

3. For 二次方程px^2+2qx-p+2p=0的根是有理數, discriminant not less than 0
Discriminant = (2q)^2 - 4(p)(-p+2p)
= 4q^2 - 4p^2
sorry, I can&#39;t prove it.
but if the question is 若p和q均為有理數，且p不等於0，證明二次方程px^2+2qx-p+2q=0的根是有理數。
For 二次方程px^2+2qx-p+2q=0的根是有理數, discriminant not less than 0
Discriminant = (2q)^2 - 4(p)(-p+2q)
= 4q^2 + 4p^2 - 8pq
= 4(p - q)^2 which is not less than zero
Therefore 二次方程px^2+2qx-p+2q=0的根是有理數.

4. For (m^2+n^2)x^2-2(m+n)x+2=0有非實數根, discriminant smaller than 0
Discriminant = [-2(m+n)]^2 - 4(m^2+n^2)(2)
=4m^2 + 8mn + 4n^2 - 8m^2 - 8n^2
= -(4m^2 - 8mn + 4n^2)
= -4(m - n)^2 which is smaller than 0 (m=/=n)(given)
Therefore 方程(m^2+n^2)x^2-2(m+n)x+2=0有非實數根.

5. (x-a)(x-b)=c
x^2 -(a+b)x + ab - c = 0
For 方程(x-a)(x-b)=c的根是實數, discriminant not smaller than 0
Discriminant = [-(a+b)]^2 - 4(ab - c)
= a^2 + 2ab + b^2 - 4ab + 4c
= a^2 - 2ab + b^2 + 4c
= (a-b)^2 + 4c which is not smaller than zero (c bigger than 0)(given)
Therefore 方程(x-a)(x-b)=c的根是實數.