Question

answer with explanation otherwise reported​

Answer 1

Answer:

Plzz find the mistake TT

we will solve together

Step-by-step explanation:

$\sqrt{2}^{2x^{2}+4x } =\frac{1}{2}$

$\sqrt{2}^{2(x^{2}+2x)} =\frac{1}{2}$

$(2^{\frac{1}{2} }) ^{2(x^{2}+2x)} =\frac{1}{2}$

$2^{x^{2}+2x} =\frac{1}{2}$

$2^{x^{2}+2x} =2^{-1}$

equating powers,

$x^{2} +2x=-1$

I tried TT

Answer 2

$\star\:\:\:\sf\large\underline\blue{Question:-}$

• Solve the given equation.

$\star\:\:\:\sf\large\underline\blue{Solution:-}$

Given that,

$\sf { (\sqrt{2} )}^{2 {x}^{2} + 4x} = \frac{1}{2}$

$\sf \implies {2}^{ \frac{1}{2} (2 {x}^{2} + 4x) } = {2}^{ - 1}$

$\sf \implies {2}^{ {x}^{2} + 2x} = {2}^{ - 1}$

Comparing base, we get,

$\sf{x}^{2} + 2x = - 1$

$\sf \implies {x}^{2} + 2x + 1 = 0$

$\sf \implies {(x + 1)}^{2} = 0$

$\sf \implies (x + 1)(x + 1)= 0$

Therefore, either x+1=0 or x+1=0

So,

$\sf x = - 1 \: and \: x = - 1$

There are two roots for the given equation. Both are same.

While solving this kind of problems, we have to make the bases equal and then comparing base, we will solve the equation.