Question

solve the following question​

Answer 1

Answer:

hope it will help you dear

Answer 2

Given Question :-

Solve for x :-

$\rm :\longmapsto\: {2}^{2x} + 3. {2}^{x} - 4 = 0$

$\red{\large\underline{\sf{Solution-}}}$

Consider the equation,

$\rm :\longmapsto\: {2}^{2x} + 3. {2}^{x} - 4 = 0$

can be rewritten as

$\rm :\longmapsto\: {( {2}^{x}) }^{2} + 3. {2}^{x} - 4 = 0$

Let we assume that

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: {2}^{x} \: = \: y \: }}}$

So, above equation can be rewritten as

$\rm :\longmapsto\: {y}^{2} + 3y - 4 = 0$

$\rm :\longmapsto\: {y}^{2} + 4y - y - 4 = 0$

$\rm :\longmapsto\:y(y + 4) - 1(y + 4) = 0$

$\rm :\longmapsto\:(y + 4)(y - 1) = 0$

$\bf\implies \:y = 1 \: \: or \: \: y = - 4$

$\bf\implies \: {2}^{x} = 1 \: \: or \: \: {2}^{x} = - 4 \: \{rejected \}$

$\bf\implies \: {2}^{x} = {2}^{0} \:$

$\bf\implies \: x = 0 \:$

Verification :-

Given equation is

$\rm :\longmapsto\: {2}^{2x} + 3. {2}^{x} - 4 = 0$

On substituting x = 0, we get

$\rm :\longmapsto\: {2}^{2 \times 0} + 3. {2}^{0} - 4 = 0$

$\rm :\longmapsto\: {2}^{0} + 3 \times 1 - 4 = 0$

$\rm :\longmapsto\: 1 + 3 - 4 = 0$

$\rm :\longmapsto\: 4 - 4 = 0$

$\bf\implies \:0 = 0$

Hence, Verified