Question

If 15a²+4b²/15a²-4b²=47/7 then find the values of b²-2a²/b²+2a²

Answer 1

Answer:

$\rm \dfrac{b^2-2a^2}{b^2+2a^2} = \dfrac 7{43}.$

Step-by-step explanation:

Given:

$\rm \dfrac{15a^2+4b^2}{15a^2-4b^2}=\dfrac {47}{7}.$

According to Componendo and Dividendo rule, if there is a fraction such as,  

$\rm \dfrac ND=\dfrac mn,$

then,

$\rm \dfrac{N+D}{N-D}=\dfrac{m+n}{m-n}.$

On applying the rule to the given equation by taking,

$\rm N= 15a^2+4b^2\\D=15a^2-4b^2\\m=47\\n=7$

we get,

$\rm \dfrac{(15a^2+4b^2)+(15a^2-4b^2)}{(15a^2+4b^2)-(15a^2-4b^2)}=\dfrac{47+7}{47-7}\\\dfrac{15a^2+15a^2+4b^2-4b^2}{15a^2-15a^2+4b^2-(-4b^2)}=\dfrac{54}{40}\\\dfrac{30a^2}{8b^2} =\dfrac{27}{20}\\\dfrac{a^2}{b^2} = \dfrac {8}{30} \times \dfrac{27}{20} = \dfrac {9}{25}\\\Rightarrow \dfrac ab=\pm \sqrt{\dfrac{9}{25}} = \pm \dfrac 35\\\Rightarrow a=\pm \dfrac 35 b.$

$\rm a^2$ is always positive, therefore there is only one value of $\rm \dfrac{b^2-2a^2}{b^2+2a^2}$, which is given by,

$\rm \dfrac{b^2-2a^2}{b^2+2a^2}=\dfrac{b^2-2\left (\dfrac 35b \right )^2}{b^2+2\left (\dfrac 35b \right )^2}\\=\dfrac{b^2-2\left (\dfrac 9{25} b^2\right )}{b^2+2\left (\dfrac 9{25} b^2\right )}\\=\dfrac{b^2-\dfrac {18}{25} b^2}{b^2+\dfrac {18}{25} b^2}\\=\dfrac{25b^2-18 b^2}{25b^2+18 b^2}\\=\dfrac{7b^2}{43b^2}\\=\dfrac{7}{43}.$