MHB 5.t.11 find x for the imaginary factors

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The discussion focuses on solving the equation f(x)=0 with roots 5+i and 5-i, highlighting errors in the original calculations. It points out that the function was incorrectly stated as quadratic when it should be cubic due to the missing factor (x-1). The correct expansion of the quadratic should yield a constant term of 24 instead of 26, impacting the quadratic formula's results. Additionally, the coefficient b was misidentified, leading to further calculation errors. Overall, the conversation emphasizes the importance of accurately stating the function and carefully executing algebraic operations.
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$\textbf{5.t.11 }$ McKinley HS

Find x for $f(x)=0 \quad 5+i\quad 5-i\quad $
$\begin{array}{rl}
\textsf{factored} &f(x)=(x-1)[x-(5+i))(x-(5-i)]\\
\textsf{foil} &x^2-x(5+i)-x(5-i)+(5-i)^2\\
\textsf{expand} &x^2-5x-xi-5x+xi+25-2i+i^2 \\
\textsf{simplify} &x^2-10x+26\\
\textsf{observation } &(x-1)=0,\quad x=1\\
\textsf{quadratic formula} &=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\
&=\dfrac{-(10)\pm\sqrt{(10)^2-4(1)(26)}}{2(1)}\\
&=\dfrac{-1\pm\sqrt{100-96}}{2}
=\dfrac{-10\pm2i}{2}=5\pm i
\end{array}$

can't seem to get the errors out of this;)
 
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karush said:
$\textbf{5.t.11 }$ McKinley HS

Find x for $f(x)=0 \quad 5+i\quad 5-i\quad $
$\begin{array}{rl}
\textsf{factored} &f(x)=(x-1)[x-(5+i))(x-(5-i)]\\
\textsf{foil} &x^2-x(5+i)-x(5-i)+(5-i)^2\\
\textsf{expand} &x^2-5x-xi-5x+xi+25-2i+i^2 \\
\textsf{simplify} &x^2-10x+26\\
\textsf{observation } &(x-1)=0,\quad x=1\\
\textsf{quadratic formula} &=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\
&=\dfrac{-(10)\pm\sqrt{(10)^2-4(1)(26)}}{2(1)}\\
&=\dfrac{-1\pm\sqrt{100-96}}{2}
=\dfrac{-10\pm2i}{2}=5\pm i
\end{array}$

can't seem to get the errors out of this;)
First: Your problem didn't state that f(1) = 0.

Second: [math]f(x)=(x-1)[x-(5+i))(x-(5-i))][/math] gives the zeros 1, 5 - i, and 5 + i but is a cubic. You left out the (x - 1) factor and got a quadratic. You never stated f(x).

Third: The last term in the quadratic expansion is [math](5 + i)(5 - i) = 25 - i^2 = 26[/math]

(Fourth: 25 + i^2 = 25 - 1 = 24. Your wrote 26 in the quadratic formula, which is correct but your work would have set c = 24 and given the wrong answer.)

Fifth: b = -10, not b = 10.

Sixth: [math]100 - 4 \cdot 26 = -4[/math], not 4.

You need to drink more coffee when you are doing these.

-Dan
 
ok thanks
 
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