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

• MHB
• karush
In summary: I have revised my solution and the correct summary is:In summary, we are trying to find the value of x for the function f(x)=0, with the given factors of (x-1), (x-(5+i)), and (x-(5-i)). By expanding and simplifying, we get the quadratic equation x^2-10x+26, where we observe that (x-1) is one of the roots. Using the quadratic formula, we can find the other two roots to be 5+i and 5-i. However, it was not stated in the problem that f(1)=0, so we cannot assume it as a root. Additionally, there were some errors in the
karush
Gold Member
MHB
$\textbf{5.t.11 }$ 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;)

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: $$\displaystyle f(x)=(x-1)[x-(5+i))(x-(5-i))]$$ 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

ok thanks

## 1. What does "5.t.11 find x for the imaginary factors" mean?

5.t.11 refers to a specific problem or equation that involves finding the value of x. The term "imaginary factors" refers to complex numbers or solutions that involve the square root of a negative number.

## 2. How do you solve for x in an equation with imaginary factors?

To solve for x in an equation with imaginary factors, you can use the quadratic formula or factor the equation to find the roots. The solutions will be complex numbers in the form of a + bi, where a and b are real numbers and i is the imaginary unit.

## 3. Can an equation have both real and imaginary factors?

Yes, an equation can have both real and imaginary factors. This means that the solutions will be a combination of real and complex numbers.

## 4. Is it possible to have more than one solution for x when dealing with imaginary factors?

Yes, it is possible to have more than one solution for x when dealing with imaginary factors. This is because complex numbers have two roots, known as the principal root and the conjugate root.

## 5. How are equations with imaginary factors used in real-life applications?

Equations with imaginary factors are used in a variety of fields, including engineering, physics, and economics. They can be used to model systems with oscillating or periodic behavior, such as electrical circuits, sound waves, and stock market fluctuations.

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