find the value of k for which sum of the root of the equation 3x^2-(3x-2)x-(k-6)=0 is equal to the product of its root
mysticd:
plz , verify the middle term
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In the Given equation ;
3x^2 - (3x-2)x -(k-6) = 0
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3x^2 -(3x^2-6) -k+6 = 0
3x^2 -3x^2+6-k+6 = 0
12 -k = 0
k = -12
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a = 3
b = -(3x-2)
c = -(k-6)
Sum of roots = -b/a
= -[-(3x-2)]/3
= (3x-2)/3
Product of roots = c/a
= -(k-6)/3
= (6-k)/3
Sum of roots = Product is roots
(3x-2)/3 = (6-k)/3
3x-2 = 6-k
k = 6+2-3x
k = 8-3x
________________
3x^2 - (3x-2)x -(k-6) = 0
__________________
3x^2 -(3x^2-6) -k+6 = 0
3x^2 -3x^2+6-k+6 = 0
12 -k = 0
k = -12
__________________
a = 3
b = -(3x-2)
c = -(k-6)
Sum of roots = -b/a
= -[-(3x-2)]/3
= (3x-2)/3
Product of roots = c/a
= -(k-6)/3
= (6-k)/3
Sum of roots = Product is roots
(3x-2)/3 = (6-k)/3
3x-2 = 6-k
k = 6+2-3x
k = 8-3x
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sardar , here , a = 0,we have to expand the middle term
sir, but according to question 'a' is 3
there are two x^2 terms
3x^2 -3x^2+2x-(k-6)=0
it becomes 2x-(k-6)=0
got it !
if a=0 then sum of the roots = -b/a = un defined
yeah, the question is wrong sir..
Ok thanks you
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