Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 0: Problem 8
Use the quadratic formula to solve the equations. $$x^{2}-4 x+5=0$$
Short Answer
Expert verified
\( x = 2 + i \) or \( x = 2 - i \)
Step by step solution
01
Identify coefficients
Identify the coefficients in the quadratic equation. The quadratic equation is in the form \[ ax^2 + bx + c = 0 \] For the given equation \[ x^2 - 4x + 5 = 0 \] we have: \[ a = 1, \ b = -4, \ c = 5 \]
02
Write the quadratic formula
The quadratic formula is given by \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
03
Substitute the coefficients into the formula
Substitute the values of \( a = 1 \), \( b = -4 \), and \( c = 5 \) into the quadratic formula:\[ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(5)}}{2(1)} \]
04
Simplify the equation under the square root
Simplify the expression under the square root:\[ x = \frac{4 \pm \sqrt{16 - 20}}{2} \] which simplifies to\[ x = \frac{4 \pm \sqrt{-4}}{2} \]
05
Simplify the square root
The expression under the square root is \(-4\), which is a negative number. We can write it using imaginary numbers: \[ \sqrt{-4} = 2i \] Thus, we have: \[ x = \frac{4 \pm 2i}{2} \]
06
Finalize the solution
Divide the terms by 2 to simplify:\[ x = 2 \pm i \]
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
solving quadratic equations
Quadratic equations often come in the form: \[ ax^2 + bx + c = 0 \].
An example would be \[x^2 - 4x + 5 = 0 \]. To solve such equations, the quadratic formula is very handy. The formula is: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. Here's how to use it:
An example would be \[x^2 - 4x + 5 = 0 \]. To solve such equations, the quadratic formula is very handy. The formula is: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. Here's how to use it:
- First, identify the coefficients from the quadratic equation. For \[x^2 - 4x + 5\], we have \[a = 1\], \[b = -4\], and \[c = 5\].
- Next, substitute these values into the quadratic formula. So, you get \[ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(5)}}{2(1)} \].
- Simplify the equation step by step. Under the square root sign, you simplify \[ \sqrt{16 - 20} \], which gives \[ \sqrt{-4} \]. This means the solution will include imaginary numbers because of the negative sign under the square root.
- Finally, the simplified solution is \[ x = 2 \pm i \].
imaginary numbers
In some quadratic equations, you may encounter imaginary numbers. If you see a negative number under the square root sign, that's your clue. For example, in the equation \[x^2 - 4x + 5 = 0\], after simplifying, you get \[\sqrt{-4}\].
- To handle this, you use the imaginary unit \[i\], where \[i^2 = -1\].
- So, \[ \sqrt{-4} = 2i \].
- This means your equation will have solutions involving imaginary numbers, like \[ x = 2 \pm i \].
quadratic equation coefficients
Quadratic equation coefficients are constants that determine the shape and position of the parabola represented by the equation. Let's dissect them:
- \[ a \]: This is the coefficient in front of \[x^2\]. It affects the parabola's width and direction (if \[a > 0\], it opens upwards; if \[a < 0\], it opens downwards). For \[ x^2 - 4x + 5 = 0 \], \[a = 1\].
- \[ b \]: This coefficient is in front of \[x\]. It affects the position of the vertex along the x-axis. For \[ x^2 - 4x + 5 = 0 \], \[b = -4\].
- \[ c \]: This is the constant term. It affects the position of the parabola along the y-axis. For \[ x^2 - 4x + 5 = 0 \], \[c = 5\].
