5.t.11 find x for the imaginary factors

[karush]

$\textbf{5.t.11 }$ 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;)

 

[topsquark]

$\textbf{5.t.11 }$ 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: \(\displaystyle f(x)=(x-1)[x-(5+i))(x-(5-i))]\) 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 \(\displaystyle (5 + i)(5 - i) = 25 - i^2 = 26\)

(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: \(\displaystyle 100 - 4 \cdot 26 = -4\), not 4.

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

-Dan

 

[karush]

ok thanks