急!!!F.4 a.maths problem x2 15分

1.

x^2 - ( 2m - 1 )x - 2m = 0

Discriminant

= ( 2m - 1 )^2 - 4(1)(-2m)

= 4m^2 - 4m + 8m + 1

= 4m^2 + 4m + 1

= (2m + 1)^2

Therefore root (2m + 1)^2 = 2m + 1,

which is rational when m is a rational number.

Therefore, the roots of the quadratic equation x^2 - ( 2m - 1 )x - 2m=0

are rational for all rational numbers m

( Note: The quadratic equation is [ -b +- root (b^2 - 4ac) ] / 2a.

Therefore if we show that root (b^2 - 4ac) is rational, the roots of the equation

will be rational too. )




2.

qx^2 + ( p + 3q )x + 2p = 0

Discriminant

= ( p + 3q )^2 - 4(q)(2p)

= p^2 + lpq + 9q^2 - 8pq

= 9q^2 - 2pq + p^2

= ( p^2 - 2pq + q^2 ) + 9q^2 - q^2

= ( p - q )^2 + 8q^2

( p - q )^2 >= 0

( p - q )^2 + 8q^2 >= 8q^2

Therefore, the discriminant of the equation is >= 8q^2.

When q > 0, 8q^2 > 0,

which shows that the roots of the equation are real

(1)

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Sep07/Crazyquadratic5.jpg