Question

if
$\frac{a}{b} = \frac{3}{2} then \: find \: the \: value \: of \: \frac{4 {a}^{2} + 3 {b}^{2} }{4 {a}^{2} - 3 {b}^{2} }$
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Answer 1

Answer:

• ➳ 2

GivEn:-

• $\sf\dfrac{a}{b} = \dfrac{3}{2}$

To Find :-

• $\sf value \: of \: \dfrac{4 {a}^{2} + 3 {b}^{2} }{4 {a}^{2} - 3 {b}^{2} }$

Step-by-step explanation:

➠ We will find the value with two simple methods! Let's do it.

➤ Method 1 :-

$\sf\dfrac{a}{b} = \dfrac{3}{2}$

$\implies \sf \: 2a = 3b$

Now,

$\sf \dfrac{4 {a}^{2} + 3 {b}^{2} }{4 {a}^{2} - 3 {b}^{2} }$

$\implies \sf \: \dfrac{ {(2a)}^{2} + 3 {b}^{2} }{{(2a)}^{2} - 3 {b}^{2}}$

As, we get 2a = 3b, we will put the value of 2a there!

$\implies \sf \: \dfrac{ {(3b)}^{2} + 3 {b}^{2} }{{(3b)}^{2} - 3 {b}^{2}}$

$\implies \sf \: \dfrac{ 9b^{2} + 3 {b}^{2} }{{9b}^{2} - 3 {b}^{2}}$

$\implies \sf \: \dfrac{ 12b^{2} }{6 {b}^{2} }$

$\boxed{\implies \sf \: 2}$

Now, another method!

➤ Method 2 :-

$\sf\dfrac{a}{b} = \dfrac{3}{2}$

➠ a = 3k

➠ b = 2k (where k,common factor ≠ 0)

$\sf \dfrac{4 {a}^{2} + 3 {b}^{2} }{4 {a}^{2} - 3 {b}^{2} }$

Now putting the values, we would get,

$\implies \sf \dfrac{4 \times ({3k})^{2} + 3 \times ({2k})^{2} }{4 \times ({3k})^{2} - 3 \times ({2k})^{2} }$

$\implies \sf \: \dfrac{4 \times 9 {k}^{2} + 3 \times 4 {k}^{2} }{4 \times 9 {k}^{2} - 3 \times 4 {k}^{2} }$

$\implies \sf \: \dfrac{36 {k}^{2} + 12 {k}^{2} }{36 {k}^{2} - 12{k}^{2} }$

$\implies \sf \dfrac{ {48k}^{2} }{ {24k}^{2} }$

$\boxed{\implies \sf \: 2}$

Therefore, the answer would be ➾ 2

Answer 2

We are given:-

$\sf\dfrac{a}{b} = \dfrac{3}{2}$

To Find :-

$\sf \dfrac{4 {a}^{2} + 3 {b}^{2} }{4 {a}^{2} - 3 {b}^{2} }$

Solution:-

$\sf\dfrac{a}{b} = \dfrac{3}{2}$

$\to{\underline{\boxed \sf \: 2a = 3b}}$

$\sf \dfrac{4 {a}^{2} + 3 {b}^{2} }{4 {a}^{2} - 3 {b}^{2} }$

$\to\sf \: \dfrac{ {(2a)}^{2} + 3 {b}^{2} }{{(2a)}^{2} - 3 {b}^{2}}$

As; 2a = 3b,

$\to \sf \: \dfrac{ {(3b)}^{2} + 3 {b}^{2} }{{(3b)}^{2} - 3 {b}^{2}}$

$\to \sf \: \dfrac{ 9b^{2} + 3 {b}^{2} }{{9b}^{2} - 3 {b}^{2}}$

$\to \sf \: \dfrac{ 12b^{2} }{6 {b}^{2} }$

${\underline {\boxed{\to \sf \: 2} }}$