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3 years ago

(-2x-1)^2=0 How to solve brackets, please help

2 answers:

5 0

All you have to know is
( a + b )² = ( a + b ) ( a + b )
In this case,
Since "a" is negative, make it into positive for easier simplification.

(-2x-1)² = 0
-(2x+1)² = 0
-[(2x+1) (2x+1)] = 0
Negative got eliminated when divided to 0
Thus,
x = -1/2 , x = -1/2

malfutka [58]3 years ago

3 0

Not quite sure what you mean by brackets, but you can get the solution to this equation in a few steps:

(-2x - 1)² = 0 ... square root both sides to eliminate the squared binomial
√(-2x - 1)² = √0 ... simplify; the square is canceled out and the root of 0 brings you back to 0
-2x - 1 = 0 ... solve like a two-step equation
-2x = 1
x = -1/2 is your x-value.

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Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \int\limits^1_0 {x^5e^{x^3 + 1}} \, dx$

Step 2: Integrate Pt. 1

[Integrand] Rewrite [Exponential Rule - Multiplying]: $\displaystyle \int\limits^1_0 {x^5e^{x^3 + 1}} \, dx = \int\limits^1_0 {x^5e^{x^3}e} \, dx$
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Step 3: Integrate Pt. 2

Identify variables for u-solve.

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[u] Differentiate [Basic Power Rule]: $\displaystyle du = 3x^2 \ dx$
[u] Rewrite: $\displaystyle x = \sqrt[3]{u}$
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Step 4: Integrate Pt. 3

[Integral] U-Solve: $\displaystyle \int\limits^1_0 {x^5e^{x^3 + 1}} \, dx = e\int\limits^1_0 {x^5e^{(\sqrt[3]{u})^3}\frac{1}{3x^2}} \, du$
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Step 6: Integrate Pt. 5

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