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3 years ago

Which real numbers are zeros of the function? f(x)=2x4−9x3+7x2+6x

Mathematics

2 answers:

meriva3 years ago

8 0

Answer: The real zeroes of the function f(x) are $0,~2,~3,~-\dfrac{1}{2}.$

Step-by-step explanation: We are given to find the real numbers that are zeroes of the following functions :

$f(x)=2x^4-9x^3+7x^2+6x~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

We know that

the zeroes of a polynomial function y = p(x) are given by p(x) = 0.

So, the zeroes of the given polynomial (i) are

$f(x)=0\\\\\Rightarrow 2x^4-9x^3+7x^2+6x=0\\\\\Rightarrow x(2x^3-9x^2+7x+6)=0\\\\\Rightarrow x=0,~~2x^3-9x^2+7x+6=0.$

So, one of the real zeroes of f(x) is x = 0 and the other zeroes are given by $2x^3-9x^2+7x+6=0.$

<em><u>Factor theorem :</u></em> If at x = a, the value of a function p(x) is zero, then (x-a) is a factor of p(x).

Let us consider that

$g(x)=2x^3-9x^2+7x+6.$

At x = 1, we have

$g(x)=2\times1^4-9\times1^3+7\times1^2+6\times1=2-9+7+6=6\neq0.$

At x = 2, we have

$g(x)=2\times1^3-9\times2^2+7\times2+6=8-36+14+6=0.$

So, (x-2) is a factor of g(x).

Therefore, we get

$g(x)\\\\=2x^3-9x^2+7x+6\\\\=2x^2(x-2)-5x(x-2)-3(x-2)\\\\=(x-2)(2x^2-5x-3)\\\\=(x-2)(2x^2-6x+x-3)\\\\=(x-2)(2x+1)(x-3).$

That is, the other two zeroes are given by

$2x+1=0,~~~~~x-3=0\\\\\Rightarrow x=-\dfrac{1}{2},3.$

Since all the four zeroes are real, so all the real zeroes of f(x) are $0,~2,~3,~-\dfrac{1}{2}.$

Thus, the real zeroes of the function f(x) are $0,~2,~3,~-\dfrac{1}{2}.$

11Alexandr11 [23.1K]3 years ago

3 0

-1/2, 0, 2, and 3 are the answers

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Since you mentioned calculus, perhaps you're supposed to find the area by integration.

The square is circumscribed by a circle of radius 6, so its diagonal (equal to the diameter) has length 12. The lengths of a square's side and its diagonal occur in a ratio of 1 to sqrt(2), so the square has side length 6sqrt(2). This means its sides occur on the lines $x=\pm3\sqrt2$ and $y=\pm3\sqrt2$ .

Let $R$ be the region bounded by the line $x=3\sqrt2$ and the circle $x^2+y^2=36$ (the rightmost blue region). The right side of the circle can be expressed in terms of $x$ as a function of $y$ :

$x^2+y^2=36\implies x=\sqrt{36-y^2}$

Then the area of this circular segment is

$\displaystyle\iint_R\mathrm dA=\int_{-3\sqrt2}^{3\sqrt2}\int_{3\sqrt2}^{\sqrt{36-y^2}}\,\mathrm dx\,\mathrm dy$

$=\displaystyle\int_{-3\sqrt2}^{3\sqrt2}(\sqrt{36-y^2}-3\sqrt2)\,\mathrm dy$

Substitute $y=6\sin t$ , so that $\mathrm dy=6\cos t\,\mathrm dt$

$=\displaystyle\int_{-\pi/4}^{\pi/4}6\cos t(\sqrt{36-(6\sin t)^2}-3\sqrt2)\,\mathrm dt$

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Then the area of the entire blue region is 4 times this, a total of $\boxed{36\pi-72}$ .

Alternatively, you can compute the area of $R$ in polar coordinates. The line $x=3\sqrt2$ becomes $r=3\sqrt2\sec\theta$ , while the circle is given by $r=6$ . The two curves intersect at $\theta=\pm\dfrac\pi4$ , so that

$\displaystyle\iint_R\mathrm dA=\int_{-\pi/4}^{\pi/4}\int_{3\sqrt2\sec\theta}^6r\,\mathrm dr\,\mathrm d\theta$

$=\displaystyle\frac12\int_{-\pi/4}^{\pi/4}(36-18\sec^2\theta)\,\mathrm d\theta=9\pi-18$

so again the total area would be $36\pi-72$ .

Or you can omit using calculus altogether and rely on some basic geometric facts. The region $R$ is a circular segment subtended by a central angle of $\dfrac\pi2$ radians. Then its area is

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so the total area is, once again, $36\pi-72$ .

An even simpler way is to subtract the area of the square from the area of the circle.

$\pi6^2-(6\sqrt2)^2=36\pi-72$

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