Question

If a,b,c,d are in proportion then prove
(11a+15b)/(11c+15d)=[(3a^2x+5b^2y)/(3c^2x+5d^2y)]^1/2

Answer 1

Answer:

Given,

a,b,c,d are in a proportion, that is $\frac{a}{b}=\frac{c}{d}$    ...... (1)

We have to prove : $\frac{11a+15b}{11c+15d}=(\frac{3a^2x+5b^2y}{3c^2x+5d^2y})^\frac{1}{2}$

L.H.S.

$\frac{11a+15b}{11c+15d}=\frac{b(11(\frac{a}{b})+15)}{d(11(\frac{c}{d})+15)}$

$=\frac{b(11(\frac{c}{d})+15)}{d(11(\frac{c}{d})+15)}$  (from equation (1))

$=\frac{b}{d}$

R.H.S.

$(\frac{3a^2x+5b^2y}{3c^2x+5d^2y})^\frac{1}{2}$

$=\sqrt{\frac{3a^2x+5b^2y}{3c^2x+5d^2y}}$

$=\sqrt{\frac{b^2(3(\frac{a}{b})^2x+5y)}{d^2(3(\frac{c}{d})^2x+5y)}}$

$=\sqrt{\frac{b^2(3(\frac{c}{d})^2x+5y)}{d^2(3(\frac{c}{d})^2x+5y)}}$                           (From equation (1) )

$=\sqrt{\frac{b^2}{d^2}}$

$=\frac{b}{d}$

Therefore, $\frac{11a+15b}{11c+15d}=(\frac{3a^2x+5b^2y}{3c^2x+5d^2y})^\frac{1}{2}$

#Learn more:

Given a/b=c/d; prove that :- (3a-5b)/(3a+5b)=(3c-5d)/(3c+5d)...

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