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Alenkasestr [34]
3 years ago
6

Find a polynomial with integer coefficients that satisfies the given conditions.

Mathematics
1 answer:
stepladder [879]3 years ago
8 0
\bf \textit{difference of squares}
\\ \quad \\
(a-b)(a+b) = a^2-b^2\qquad \qquad 
a^2-b^2 = (a-b)(a+b)
\\\\\\
\textit{also recall that }~~~i^2=-1\\\\
-------------------------------\\\\
\begin{cases}
x=5+i\implies &x-5-i=0\\
x=5-i\implies &x-5+i=0
\end{cases}
\\\\\\
(x-5-i)(x-5+i)=\stackrel{\textit{original polynomial}}{0}
\\\\\\\
[(x-5)~~-~~i]~[(x-5)~~+~~i]=0\implies [(x-5)^2~~-~~i^2]=0
\\\\\\
(x^2-10x+25)~~-~~(-1)=0\implies x^2-10x+25+1=y
\\\\\\
x^2-10x+26=y
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Solve for x in the equation x squared + 2 x + 1 = 17.
Papessa [141]

Answer:

x = - 1 + \sqrt{17}\\and\\x = - 1 - \sqrt{17}\\

Step-by-step explanation:

given equation

x^2 +2x +1 = 17

subtracting 17 from both sides

x^2 +2x +1 = 17\\x^2 +2x +1 -17= 17-17\\x^2 +2x - 16 = 0\\

the solution for quadratic equation

ax^2 + bx + c = 0 is given by

x = x = -b + \sqrt{b^2 - 4ac} /2a \\\\and \ \\-b - \sqrt{b^2 - 4ac} /2a

________________________________

in our problem

a = 1

b = 2

c = -16

x =( -2 + \sqrt{2^2 - 4*1*-16}) /2*1 \\x =( -2 + \sqrt{4  + 64}) /2\\x =( -2 + \sqrt{68} )/2\\x = ( -2 + \sqrt{4*17} )/2\\x =  ( -2 + 2\sqrt{17} )/2\\x = - 1 + \sqrt{17}\\and\\\\x = - 1 - \sqrt{17}\\

thus value of x is

x = - 1 + \sqrt{17}\\and\\x = - 1 - \sqrt{17}\\

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2.0Plss solve for x...........................
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Answer:

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