Question

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If (x-3) and (x-1/3) are both factors of ax^2+5x+b show that a=b

Answer 1

a x² + 5x +b............(1)

it is 2 degree equation so, it has 2 factors which are

(x-3), (x-1/3)

we can derive an equation by multiplying these 2 factors

(x-3)(x - 1/3)

x²-3x -(1/3 )x +1

= x² -10x/3 + 1.........(2)

by comparing (1) and (2) equation

we get,

a = 1

b = 1

hence ,

a = b = 1

or

a = b

Answer 2

P(x) = ax^2+5x+b

To Prove:-

a=b

x-3 & x-1/3 are factors

x=3 & x=1/3 are zeroes

put x=3 in p(x)

a(3)^2+5(3)+b=0

9a+15+b=0

9a+b=-15  ------------(1)

put x=1/3 in p(x)

a(1/3)^2+5(1/3)+b=0

a/9+5/3+b=0

a/9+b=-5/3  {taking L.C.M.}

a+9b=-5/3*9

a+9b=-15 ------------(2)

subtracting (1) & (2)

9a+b=-15

a+9b=-15

(-)  (-)     (+)             {subtracting thus changing signs}

___________

8a - 8b = 0

8a = 8b

a=8b/8

a=b

H.P.

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