Mathematics, 26.07.2021 08:30 arias333

Please, look at this one.

Please,look at this one.

Answers: 2

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Answer from: Nerdymania

9514 1404 393

x = √2

Step-by-step explanation:

A graph indicates the only solution is near x=√2.

Square both sides, separate the radical and do it again.

$\displaystyle(2-x)\sqrt{\frac{x+2}{x-1}}=\sqrt{x}+\sqrt{3x-4}\qquad\text{given}\\\\(2-x)^2\cdot\frac{x+2}{x-1}=x+(3x-4)+2\sqrt{x(3x-4)}\qquad\text{square}\\\\\frac{(2-x)^2(x+2)}{x-1}-4x+4=2\sqrt{x(3x-4)}\qquad\text{isolate radical}\\\\\left(\frac{(2-x)^2(x+2)-4(x-1)^2}{x-1}\right)^2=x(3x-4)\qquad\text{square}\\\\(x^3-6x^2+4x+4)^2=4(x-1)^2(3x^2-4x)\qquad\text{multiply by $(x-1)^2$}$

Now, we can put this polynomial equation into standard form and factor it.

$x^6 -12x^5+32x^4-76x^2+48x+16=0\\\\(x-2)^2(x^2-2)(x^2-8x-2)=0\qquad\text{factor it}\\\\x\in \{2,\pm\sqrt{2},4\pm3\sqrt{2}\}$

The original equation requires that we restrict the domain of possible solutions. In order for the radicals to be non-negative, we must have x ≥ 4/3. In order for the left side of the equation to be non-negative, we must have x ≤ 2. So, the only potential solutions will be in the interval [4/3, 2].

The only values in the above list that match this requirement are {√2, 2}. We know that the right side of the equation cannot be zero, so the value x=2 is also an extraneous solution.

The only solution is x = √2.

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Additional comment

For solving higher-degree polynomials, I like to use a graphing calculator to help me find the roots. The second attachment shows the roots of the 6th-degree polynomial. They can help us factor the equation. (There are also various machine solvers available that will show factors and roots.)

Answer from: Quest

no problobm

step-by-step explanation:

Guys the question is in spanish but i hope some of you will undertand it, ill be so glad if you coul

Answer from: Quest

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

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