F.2 mathematics-identities

1. substitute x=0 in the equation (x-2)(x-6)=(X-4)to the power 2 -6 to show that the equation is not an identity.

(x-2)(x-6) = (x-4)^2 - 6
when x = 0, L.H.S. = (0-2)(0-6) = (-2)(-6) = 12
R.H.S. = 0-4 = (-4)^2 - 6 = 10
L.H.S. 不等於 R.H.S.
so, the equation is not an identity
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find the value of the constants A and B ineach of the following identites.

2. 3(2B-3x)=-(1-2A)x+2(B+12)
put x = 0, 3(2B)= 2(B+12)
6B= 2B+24
B = 6
put x = 1, 3(2B-3)=-(1-2A)+2(B+12)
3(2(6)-3)=-(1-2A)+2(6+12)
27 =-(1-2A)+36
A = -4

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3. (8A+3B)x+2B =-7x -6
put x = 0, 2B = -6,
B = -3
put x = 1, (8A+3B)+2B =-7 -6
8A+5B = -13
8A+5(-3) = -13
A = 1/4
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4. 2x to the power 2 + Ax-3=-2(x+1)to the power 2+B
2x^2 + Ax-3=-2(x+1)^2+B
put x = 0, -3=-2(0+1)^2+B
-3=-2+B
B = -1
put x = -1, 2(-1)^2 + A(-1)-3=-2(-1+1)^2+B
2 - A-3=B
2 - A-3= -1
A = 0

1.
LHS=(x-2)(x-6)=(-2)(-6)=12
RHS=(x-4)^2=(-4)^2=16
since LHS is not equal to RHS
so: (x-2)(x-6)is not equal to(x-4)^2

2.
3(2B-3x)=-(1-2A)x+2(B+12)
6B-9x =(2B+14) +(2A-1)x
By equating coefficient of x,
-9x=(2A-1)x
-9 =2A-1
A =-4
Similarly,B= 7/2

3.
(8A+3B)x+2B =-7x -6
By equating coefficient of constant term,
2B=-6
B =-3
similarly,A=1/4

4.我估你寫錯左條式...應該係咁??
2(x^2) + Ax-3=2[(x+1)^2]+B
2(x^2) + Ax-3=2(x^2) +4x +(B+2)
***↑e個係constant term
By equating coefficient of constant term,
-3=B+2
B=-5
similarly,A=4


應該唔會有錯...但果幾條數...similarly果度你可能要寫番詳細d...