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2 years ago

What is the solution √3x +3 - 1 = x

Mathematics

1 answer:

faltersainse [42]2 years ago

3 0

Answer:

Choice A

Step-by-step explanation:

√(3x +3) - 1 = x

Taking square on both sides:

$( \sqrt{3x + 2} ) {}^{2} = (x + 1) {}^{2}$
$3x + 3 = x { }^{2} + 2x + 1$

Subtracting x²+2x + 1 from both sides:

$3x+3−(x {}^{2} +2x+1)=x {}^{2} +2x+1−(x {}^{2} +2x+1)$
$3x + 3 - x {}^{2} - 2x - 1 = 0$
$3x - 2x + 3 - 1 - x {}^{2} = 0$
$x + 2 - x {}^{2} = 0$

Rewrite as,

$-x² + x + 2= 0$

Factoring LHS of the equation:[Middle term factor]

$( - x - 1)(x - 2) = 0$

Now,

$( - x - 1) = 0 \: \: \: \: \: ...(1)$
$x - 2 = 0 \: \: \: \: \: ...(2)$

Equation 1:

$- x - 1 = 0$
$- x = 0 + 1$
$- x = 1$
$x = \boxed{ - 1}$

Equation 2:

$x - 2 = 0$
$x = 0 + 2$
$x = \boxed{2}$

Choice A is accurate.

-1 & 2

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Answer:

The bottom one: The value is the difference of 42 and the sum of 20 and 3.

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Find the critical points of the function f(x, y) = 8y2x − 8yx2 + 9xy. Determine whether they are local minima, local maxima, or

NARA [144]

Answer:

Saddle point: $(0,0)$

Local minimum: $(\frac{3}{8}, -\frac{3}{8})$

Local maxima: $(0,-\frac{9}{8})$ , $(\frac{9}{8},0)$

Step-by-step explanation:

The function is:

$f(x,y) = 8\cdot y^{2}\cdot x -8\cdot y\cdot x^{2} + 9\cdot x \cdot y$

The partial derivatives of the function are included below:

$\frac{\partial f}{\partial x} = 8\cdot y^{2}-16\cdot y\cdot x+9\cdot y$

$\frac{\partial f}{\partial x} = y \cdot (8\cdot y -16\cdot x + 9)$

$\frac{\partial f}{\partial y} = 16\cdot y \cdot x - 8 \cdot x^{2} + 9\cdot x$

$\frac{\partial f}{\partial y} = x \cdot (16\cdot y - 8\cdot x + 9)$

Local minima, local maxima and saddle points are determined by equalizing both partial derivatives to zero.

$y \cdot (8\cdot y -16\cdot x + 9) = 0$

$x \cdot (16\cdot y - 8\cdot x + 9) = 0$

It is quite evident that one point is (0,0). Another point is found by solving the following system of linear equations:

$\left \{ {{-16\cdot x + 8\cdot y=-9} \atop {-8\cdot x + 16\cdot y=-9}} \right.$

The solution of the system is (3/8, -3/8).

Let assume that y = 0, the nonlinear system is reduced to a sole expression:

$x\cdot (-8\cdot x + 9) = 0$

Another solution is (9/8,0).

Now, let consider that x = 0, the nonlinear system is now reduced to this:

$y\cdot (8\cdot y+9) = 0$

Another solution is (0, -9/8).

The next step is to determine whether point is a local maximum, a local minimum or a saddle point. The second derivative test:

$H = \frac{\partial^{2} f}{\partial x^{2}} \cdot \frac{\partial^{2} f}{\partial y^{2}} - \frac{\partial^{2} f}{\partial x \partial y}$

The second derivatives of the function are:

$\frac{\partial^{2} f}{\partial x^{2}} = 0$

$\frac{\partial^{2} f}{\partial y^{2}} = 0$

$\frac{\partial^{2} f}{\partial x \partial y} = 16\cdot y -16\cdot x + 9$

Then, the expression is simplified to this and each point is tested:

$H = -16\cdot y +16\cdot x -9$

S1: (0,0)

$H = -9$ (Saddle Point)

S2: (3/8,-3/8)

$H = 3$ (Local maximum or minimum)

S3: (9/8, 0)

$H = 9$ (Local maximum or minimum)

S4: (0, - 9/8)

$H = 9$ (Local maximum or minimum)

Unfortunately, the second derivative test associated with the function does offer an effective method to distinguish between local maximum and local minimums. A more direct approach is used to make a fair classification:

S2: (3/8,-3/8)

$f(\frac{3}{8} ,-\frac{3}{8} ) = - \frac{27}{64}$ (Local minimum)