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If b^2-4ac=0, then find domain of y = log (ax^3 + (a + b)x^2 + (b + c)x + c) .
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b² - 4ac = 0 (given)
y = log(ax³ + (a+b)x² + (b+c)x + c)
we know that for log k , k >0
∴ (ax³ + bx² +cx )+ (ax² + bx + c) > 0
⇒ (ax² + bx + c)x + (ax² + bx + c) > 0
⇒ (ax² + bx + c) (x + 1) > 0
⇒ x > - 1 and
⇒ (ax² + bx + c) > 0
we have b² - 4ac = 0 which means the quadratic equation has real and equal solutions
which gives
⇒ x > -b/2a {a ≠ 0}
plotting the graph of both the values of x and analyzing by wave method we get x ∈ (-1 , ∞)
which is the required domain
y = log(ax³ + (a+b)x² + (b+c)x + c)
we know that for log k , k >0
∴ (ax³ + bx² +cx )+ (ax² + bx + c) > 0
⇒ (ax² + bx + c)x + (ax² + bx + c) > 0
⇒ (ax² + bx + c) (x + 1) > 0
⇒ x > - 1 and
⇒ (ax² + bx + c) > 0
we have b² - 4ac = 0 which means the quadratic equation has real and equal solutions
which gives
⇒ x > -b/2a {a ≠ 0}
plotting the graph of both the values of x and analyzing by wave method we get x ∈ (-1 , ∞)
which is the required domain
hpgr1999:
thanks for your argument about a not equal to 0
but in (ax² + bx + c) (x + 1) when both term -ve which means -ve * -ve = +ve
then log is defined
???
so whats the difference if just make it positive like that ??
u multiply two minus signs to make it positive , when it is alreast positive u r just thinking it way too hard .
already*
ohk
thanks for the solution
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