Find All Complex Solutions t^2-7t+16=t
This problem is asking for all the complex numbers that satisfy the equation t^2 - 7t + 16 = t. To find the complex solutions, one would typically move all terms to one side of the equation to set it equal to zero—thus, the corrected equation would be t^2 - 8t + 16 = 0. The task then is to solve this quadratic equation for t. The solutions could be real numbers or complex numbers, depending on the discriminant (the part of the quadratic formula under the square root). If the discriminant is negative, there will be two complex solutions. If it is zero, there will be one real solution (which is also a complex number with an imaginary part of 0), and if it is positive, there will be two distinct real solutions (which also count as complex numbers with imaginary parts of 0). Solving the equation involves factoring, using the quadratic formula, or completing the square.
$t^{2} - 7 t + 16 = t$
Answer
Solution:
Step:1
Rearrange the equation to set it to zero.
Step:1.1
Subtract $t$ from each side to get $t^{2} - 7t + 16 - t = 0$.
Step:1.2
Combine like terms to form $t^{2} - 8t + 16 = 0$.
Step:2
Apply the quadratic formula to determine the roots of the equation: $\frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$.
Step:3
Insert $a = 1$, $b = -8$, and $c = 16$ into the quadratic formula and compute for $t$: $\frac{-(-8) \pm \sqrt{(-8)^{2} - 4 \cdot 1 \cdot 16}}{2 \cdot 1}$.
Step:4
Simplify the expression.
Step:4.1
Simplify inside the square root and the numerator.
Step:4.1.1
Square $-8$ to get $t = \frac{8 \pm \sqrt{64 - 4 \cdot 1 \cdot 16}}{2 \cdot 1}$.
Step:4.1.2
Calculate the product inside the square root.
Step:4.1.2.1
Multiply $-4$ and $1$ to get $t = \frac{8 \pm \sqrt{64 - 4 \cdot 16}}{2 \cdot 1}$.
Step:4.1.2.2
Multiply $-4$ and $16$ to obtain $t = \frac{8 \pm \sqrt{64 - 64}}{2 \cdot 1}$.
Step:4.1.3
Subtract within the square root to get $t = \frac{8 \pm \sqrt{0}}{2 \cdot 1}$.
Step:4.1.4
Express $0$ as a square, resulting in $t = \frac{8 \pm \sqrt{0^{2}}}{2 \cdot 1}$.
Step:4.1.5
Extract terms from under the radical, assuming they represent positive real numbers: $t = \frac{8 \pm 0}{2 \cdot 1}$.
Step:4.1.6
Combine $8$ with $\pm 0$ to simplify to $t = \frac{8}{2 \cdot 1}$.
Step:4.2
Multiply the denominator: $t = \frac{8}{2}$.
Step:4.3
Divide the numerator by the denominator to find $t = 4$.
Step:5
Combine the solutions to present the final answer: $t = 4$, which is a repeated root.
Knowledge Notes:
The problem involves solving a quadratic equation, which is an equation of the form $ax^2 + bx + c = 0$. To solve such equations, one can use the quadratic formula: $\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. This formula provides the solutions for $x$ by substituting the coefficients $a$, $b$, and $c$ from the equation.
The process includes:
Rearranging the equation to set it equal to zero.
Identifying the coefficients $a$, $b$, and $c$.
Substituting these coefficients into the quadratic formula.
Simplifying the expression, including the square root and the fraction.
Finding the numerical solution(s), which may be real or complex numbers.
In this case, the solution is a real number, and it is a repeated root, meaning the quadratic has a perfect square trinomial, which factors into $(t - r)^2$, where $r$ is the repeated root.
