問一元二次方程式 - 10分 (只限今日)

1) 用十字相乘法,
x + a
x - b
-------------------------
ax - bx = ( a - b )x
所以x^2 + ( a - b )x - ab = 0
= > ( x + a )( x - b ) = 0
x = -a 或 b
2) 用十字相乘法,
x + 1/a
x + a
-----------------
x/a + ax = ( a + 1 / a )x
所以x^2 + ( a + 1 / a )x + 1 = 0
=> ( x + 1 / a )( x + a ) = 0
x = - 1 / a 或 -a
3) x^2 - 5x - 24 = 0
( x + 3 )( x - 8 ) = 0 [用十字相乘法]
x = -3 或8
4) 用十字相乘法,
ax - 1
x - a
-------------
-a - a^2x =- ( a^2 + 1 )x
所以ax^2 - (a^2+1)x + a = 0
=> ( ax - 1 )( x - a ) = 0
x = 1 / a 或 a
5) mx^2 = nx/p
mpx^2 - nx = 0
x ( mpx - n ) = 0
x = 0 或 n / mp
6) 用十字相乘法,
( b - c )x - ( a - b )
x - 1
-----------------------
-( a - b )x - ( b - c )x = ( c - a )x
所以(b-c)x^2 + (c - a)x + (a - b) = 0
=> [ ( b - c )x - ( a - b )]( x - 1 ) = 0
x = ( a - b ) / ( b - c ) 或 1
7) 是(p^2 - q^2) (x^2 + 1) = 2 (p^2 - q^2)x嗎?
( p^2 - q^2 )x^2 - 2 ( p^2 - q^2 )x + ( p^2 - q^2 ) = 0
用十字相乘法,
( p^2 - q^2 )x - ( p^2 - q^2 )
x - 1
-----------------------------------------
-( p^2 - q^2 )x - ( p^2 - q^2 )x = -2 ( p^2 - q^2 )x
於是[ ( p^2 - q^2 ) x - ( p^2 - q^2 )]( x - 1 ) = 0
x = 1
8) x^2y - 8 = (y^2 + 1)x
( - 2 )^2 y - 8 = ( y^2 + 1 )( - 2 )
4y - 8 = -2y^2 - 2
2y^2 + 4y - 6 = 0
y^2 + 2y - 3 = 0
( y + 3 )( y - 1 ) = 0
y = -3 或 1
9) x (x - 1) = 2
x^2 - x - 2 = 0
( x - 2 )( x + 1 ) = 0
x = 2 或 -1
(x^2 - 4 )/ sqr( x^2 - 2x + 1 )
= ( x + 2 )( x - 2 ) / sqr ( x - 1 )^2
= ( x + 2 )( x - 2 ) / ( x - 1 )
當x= 2,
( 2 + 2 )( 2 - 2 ) / ( 2 - 1 )
= 0
當x = -1,
( - 1 + 2 )( - 1 - 2 ) / ( - 1 - 1 )
= 3/2