What is the equation of a quadratic function P with rational coefficients that

Mathematics

1 answer:

Phoenix [80]4 years ago

4 0

Answer:

$P(x) = x^2-6x+58$

Step-by-step explanation:

Quadratic function is function of the form $P(x)=ax^2+bx+c$ where a, b and c are real numbers and $a\neq 0$

A number is said to be a rational number if it can be written in the form $\frac{u}{v}$ where u and v are integers and $v\neq 0$ .

A number is said to be a complex number if it can be written in the form u+iv where u and v are real numbers.

A number 'a' is said to be a zero of a function P(x) if $P(a)=0$

We know that if zero of a fraction is of form u+iv then its other zero is u-iv.

As 3+7i is a zero of the function, 3-7i is also its zero.

So, given equation is $\left ( x-(3-7i) \right )\left ( x-(3+7i) \right )$

$P(x)=\left ( x-(3-7i) \right )\left ( x-(3+7i) \right )\\=x^2-x(3+7i)-x(3-7i)+(3-7i)(3+7i)\\=x^2-3x-7ix-3x+7ix+9+49\\=x^2-6x+58$

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Write inequalities to describe the solid left (x < 0 is left) hemisphere of a sphere of radius 4 centered at the origin

Serggg [28]

The equation of the sphere centered at 0, and radius 4 is:

$\displaystyle{ x^2+y^2+z^2=4^2$ ,

note that this equation describes exactly the points of the surface of the square. That is, this is an EMPTY sphere.

The solid sphere, that is the points on the surface and all points in the inside, are given by :

$\displaystyle{ x^2+y^2+z^2 \leq 4^2$

since we want the left part of the solid part, picture 2, we add the condition x<0,

thus "the solid left (x < 0 is left) hemisphere of a sphere of radius 4 centered at the origin" is given by the system of inequalities:

$i) \displaystyle{ x^2+y^2+z^2 \leq 4^2$
$ii)x\ \textless \ 0$