Math, asked by hpgr1999, 1 year ago

Any Mathematician on Brainly ??????? (80 POINTS!!!!!)
If b^2-4ac=0, then find domain of y = log (ax^3 + (a + b)x^2 + (b + c)x + c) .

Answers

Answered by Anonymous

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

hpgr1999: thanks for your argument about a not equal to 0

hpgr1999: but in (ax² + bx + c) (x + 1) when both term -ve which means -ve * -ve = +ve

hpgr1999: then log is defined

hpgr1999: ???

Anonymous: so whats the difference if just make it positive like that ??

Anonymous: u multiply two minus signs to make it positive , when it is alreast positive u r just thinking it way too hard .

Anonymous: already*

hpgr1999: ohk

hpgr1999: thanks for the solution

Answered by Anonymous

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Any Mathematician on Brainly ??????? (80 POINTS!!!!!) If b^2-4ac=0, then find domain of y = log (ax^3 + (a + b)x^2 + (b + c)x + c) .

Answers

Any Mathematician on Brainly ??????? (80 POINTS!!!!!)
If b^2-4ac=0, then find domain of y = log (ax^3 + (a + b)x^2 + (b + c)x + c) .