續二次方程(35-37題)

35a
(1-x)(x+5)=-10x-r
x+5-x^2-5x=-10x-r
-x^2-4x+5=-10x-r
-x^2+6x+5+r=0
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(6)^2-4(-1)(5+r)<0
36+4(5+r)<0
4(5+r) < -36
5+r < -9
r < -14 //

35(b)
r的最大整數值= -15
-x^2+6x+5+(-15)=0
-x^2+6x-10=0
x^2-6x+10=0
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x= -(-6)±√[(-6)^2-4(1)(10)] / 2(1)
=6±√(36-40) / 2
=6±√(-4) / 2
=6±2i / 2
=3±i
.'.x=3+i or 3-i

36
x^2-18/x^2=7
x^4-18=7x^2
x^4-7x^2-18=0
(x^2-9)(x^2+2)=0
.'.x^2=9 or x^2=-2
i.e. x= -3 or 3 or √2 i or -√2 i

37
(x^2+4x)(x^2+4x+13)=-40
(x^2+4x)^2+13(x^2+4x)+40=0
Let a be x^2+4x
a^2+13a+40=0
(a+5)(a+8)=0
(x^2+4x+5)(x^2+4x+8)=0
x=-4±√[(4)^2-4(1)(5)] / 2(1) or -4±√[(4)^2-4(1)(8)] / 2(1)
=-4±√(16-20) / 2 or -4±√(16-32) / 2
=-4±√(-4) / 2 or -4±√(-16) / 2
=-4±2i / 2 or -4±4i / 2
=-2±i or -2±2i
.'.x= -2+i or -2-i or -2+2i or -2-2i

35) a) (1 – x)(x + 5) = - 10x– r - x² - 4x + 5 = - 10x – rx² - 6x –(r + 5) = 0Δ = 6² + 4(r + 5) < 04r < - 56 r < - 14 b) x² + 4x –(- 15 + 5) = 0x² + 4x+ 10 = 0x = [- 4 ± √(4² - 4 × 10)]/2= [- 4 ± √(- 24)]/2= - 2 ± √6 i 36) x² - 18/x² = 7x² - 7 –18/x² = 0(x – 9/x)(x + 2/x) = 0 x – 9/x = 0 or x + 2/x = 0 x² = 9or x² = - 2x = ± 3 or x = ± √2 i 37) (x² + 4x)(x² + 4x + 13) = - 40y(y + 13) = - 40 <-(Put y = x² + 4x)y² + 13y+ 40 = 0(y + 8)(y + 5) = 0 y = - 8 or y = - 5 x² + 4x + 8 = 0 or x² + 4x + 5 = 0x = [- 4 ± √(4² - 4 × 8)]/2 orx = [- 4 ± √(4² - 4 × 5)]/2x = [- 4 ± √(-16)]/2 or x = [- 4 ± √(- 4)]/2x = - 2 ± 2i or x = - 2 ± i I HOPE THIS CAN HELP YOU! ~ ^_^