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Integration to Solve Modified Sum of Binomial Coefficients

a x 2+b x+c p...

Question

$\frac{a x^{2} + b x + c}{p x^{2} + q x + r}$

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Solution

$Let u = a x^{2} + b x + c; v = p x^{2} + q x + r Then, u' = 2 a x + b; v' = 2 p x + q Using the quotient rule: \frac{d}{d x} (\frac{u}{v}) = \frac{v u' - u v'}{v^{2}} \frac{d}{d x} (\frac{a x^{2} + b x + c}{p x^{2} + q x + r}) = \frac{(p x^{2} + q x + r) (2 a x + b) - (a x^{2} + b x + c) (2 p x + q)}{{(p x^{2} + q x + r)}^{2}} = \frac{2 a p x^{3} + 2 a q x^{2} + 2 a r x + b p x^{2} + b q x + b r - 2 a p x^{3} - 2 b p x^{2} - 2 p c x - a q x^{2} - b q x - c q}{{(p x^{2} + q x + r)}^{2}} = \frac{(a q - b p) x^{2} + 2 (a r - x p) x + b r - c q}{{(p x^{2} + q x + r)}^{2}}$

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Similar questions

Differentiate the following functions with respect to x :

$\frac{a x^{2} + b x + c}{p x^{2} + q x + r}$

P (x) = {a x}^{2} + b x + c

Q (x) = {- a x}^{2} + d x + c

a c \neq 0

P (x), Q (x) = 0

P (x) = a x^{2} + a x^{2} + b x + c

Q (x) = - a x^{2} + d x + c

a c \neq 0

then P(x) Q(x)=0 has at least two real roots.

Q. The ratio of the roots of the equation

{a x}^{2} + b x + c = 0

is same as the ratio of the roots of equation

{p x}^{2} + q x + r = 0

D_{1}

D_{2}

are the discriminants of

{a x}^{2} + b x + c = 0

{p x}^{2} + q x + r = 0

respectively, then

D_{1} : D_{2} =

Q. The ratio of the roots of the equation

a x^{2} + b x + c = 0

is same as the ratio of the roots of the equation

p x^{2} + q x + r = 0

D_{1}

D_{2}

are the discriminants of

a x^{2} + b x + c = 0

p x^{2} + q x + r = 0

respectively. Find the ratio of

D_{1}

D_{2}

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