Question

answer with explanation otherwise reported​

Answer 1

Answer:

2^1/2(2x^2+4x)= 2^-1

Bases are equal now equating powers

1/2 ( 2x^2 +4x)= -1

2x^2 +4x= -2

2(x^2+ 2x) = -2

X^2+2x+1 =0

(X+1)^2=0

X+1 =0

X= -1

Step-by-step explanation:

Answer 2

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• 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.