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3 years ago

1/b+x=3b/2x^2 - 1/xbtw ^2 means square

Mathematics

1 answer:

AlekseyPX3 years ago

8 0

Answer:

$x = \frac{ - 3b}{4} \: or \: b$

Step-by-step explanation:

$\frac{1}{b + x} = \frac{3b}{2 {x}^{2} } - \frac{1}{x}$

$= > \frac{1}{b + x} = \frac{3b - 2x}{2 {x}^{2} }$

$= > 2 {x}^{2} = (b + x)(3b - 2x)$

$= > 2 {x}^{2} = 3 {b}^{2} - 2bx + 3bx - 2 {x}^{2}$

$= > 2 {b}^{2} = 3 {b}^{2} + bx - 2 {x}^{2}$

$= > 3 {b}^{2} + bx - 4 {x}^{2} = 0$

$= > 3 {b}^{2} + 4bx - 3bx - 4 {x}^{2} = 0$

$= > b(3b + 4x) - x(3b + 4x) = 0$

$= > (3b + 4x)(b - x) = 0$

Hence,

$x = \frac{ - 3b}{4} \: or \: b$

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Step-by-step explanation:

Given the two inequalities:

$y \ge 2x + 1\\ y \le 2x - 2$

To graph them, first of all, let us write their corresponding equations:

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We now find at least 2 points each which satisfy the equations to plot the graph.

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Two points are (0, 1) and ( $-\frac{1}2$ , 0).

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Putting x = 0, y = -2

Putting y = 0, x = 1.

Two points are (0, -2) and (1, 0).

Now, let us plot them.

Let us take a point (1, 4) and check whether it satisfies the first inequality.

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Which is true.

So, The shaded region will be towards point (1,4).

Let us take a point (1, 4) and check whether it satisfies the second inequality.

$4 \le 2 \times 1 -2\\\Rightarrow 4 \le 0\ [\bold{False}]$

So, The shaded region will be opposite to point (1,4).

Please refer to the attached graph for the answer.

No solution exists for them because there is no common shaded region.

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4 years ago

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3 years ago

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4 years ago

1/b+x=3b/2x^2 - 1/xbtw ^2 means square​

1/b+x=3b/2x^2 - 1/xbtw ^2 means square