<img src="https://tex.z-dn.net/?f=%5Csf%202x%2B21%3D3" id="TexFormula1" title="\sf 2x+21=3" alt="\sf 2x+21=3" align="absmiddle"

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Charra [1.4K]

3 years ago

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Mathematics

2 answers:

sergij07 [2.7K]3 years ago

5 0

$\sf \longmapsto \: 2x + 21 = 3$

$\sf \longmapsto2x = 3 - 21$

$\sf \longmapsto2x = - 18$

$\sf \longmapsto \: x = \frac{ - 18}{2}$

$\sf \longmapsto x = - 9$

grigory [225]3 years ago

3 0

Answer:

$\sf 2x+21=3$

$\sf 2x+21-21=3-21$

$\sf 2x=-18$

$\sf \cfrac{2x}{2}=\cfrac{-18}{2}$

$\boxed{\sf x=-9}$

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Consider the polynomials p(x) = 3x + 27x^2 and q(x)= 2 . Find the x -coordinate(s) of the point(s) of intersection of these two

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The x -coordinate(s) of the point(s) of intersection of these two polynomials are $x=\frac{2}{9}\approx0.2222,\:x=-\frac{1}{3}\approx-0.3333$

The sum of these x -coordinates is $\frac{2}{9}+\left(-\frac{1}{3}\right)=-\frac{1}{9}$

Step-by-step explanation:

The intersections of the two polynomials, p(x) and q(x), are the roots of the equation p(x) = q(x).

Thus, $3x + 27x^2=2$ and we solve for x

$3x+27x^2-2=2-2\\27x^2+3x-2=0\\\left(27x^2-6x\right)+\left(9x-2\right)\\3x\left(9x-2\right)+\left(9x-2\right)\\\left(9x-2\right)\left(3x+1\right)=0$

Using Zero Factor Theorem: = 0 if and only if = 0 or = 0

$9x-2=0\\9x=2\\x=\frac{2}{9}$

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The solutions are:

$x=\frac{2}{9}\approx0.2222,\:x=-\frac{1}{3}\approx-0.3333$

The sum of these x -coordinates is

$\frac{2}{9}+\left(-\frac{1}{3}\right)=-\frac{1}{9}$

We can check our work with the graph of the two polynomials.

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