( a)
Solve the equation 2(x^2) - 5x + 4=0
( b)
Hence solve the equation 2( x+1)^2 - 5x -1 = 0
quadratic equations
2009-10-15 7:26 am
回答 (3)
2009-10-15 8:09 am
✔ 最佳答案
( a) Solve the equation 2(x^2)－5x + 4=0
Sol
2x^2－5x+4=0
x=(5+/-sqrt(25－4*2*4))/4=(5+/-isqrt(7))/4
( b)Hence solve the equation 2( x+1)^2－5x－1 = 0
Sol
2( x+1)^2－5x－1 = 0
2(x+1)^2－5x－5+4=0
2(x+1)^2－5(x+1)+4=0
So
x+1=(5+/-isqrt(7))/4
x=(1+/-isqrt(7))/4
2009-10-15 8:02 am
2x^2-5x+4=0
=>(2x-1)(x-2)=0
=>x=1/2 or x=2//
b) 2(x+1)^2-5x-1=0
=>2(x+1)^2-5(x+1)+4=0
by a ) x+1 =1/2 or x+1=2
=>x=-1/2 or 1//
=>(2x-1)(x-2)=0
=>x=1/2 or x=2//
b) 2(x+1)^2-5x-1=0
=>2(x+1)^2-5(x+1)+4=0
by a ) x+1 =1/2 or x+1=2
=>x=-1/2 or 1//
2009-10-15 7:36 am
Anything wrong?
For (a), discriminant = - 7, no real roots.
For (a), discriminant = - 7, no real roots.
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