188k views
1 vote
What are the solutions to x^2-2x+5=0

User Ewerybody
by
9.4k points

2 Answers

5 votes

Answer:

x = 1 ± 2i

Explanation:

x² - 2x + 5 = 0 ( add 5 to both sides )

x² - 2x = - 5

Using the method of completing the square

add ( half the coefficient of the x- term)² to both sides

x² + 2(- 1)x + 1 = - 5 + 1

(x - 1)² = - 4 ( take square root of both sides )

x - 1 = ±
√(-4) = ± 2i ( add 1 to both sides )

x = 1 ± 2i ( that is the solutions are complex )

User Aidan
by
8.3k points
4 votes

Answer:


x=1\pm2i

or


x=1+2i


x=1-2i

Explanation:

All equations of the form
ax^2+bx+c=0 can be solved using the quadratic formula:
x = (-b \pm √(b^2-4ac))/(2a)

Quadratic formula gives two solutions, one when ± is addition add one when it is subtraction.


x^2-2x+5=0

Equation is in standard form:
ax^2+bx+c=0

Substitute 1 from a,
-2 for b, and
5 for C in the quadratic formula,
(-b\pm √(b^2-4ac))/(2a)


x=(-\left(-2\right)\pm √(\left(-2\right)^2-4* \:1* \:5))/(2)

Multiply -4 × 5:


x=(-\left(-2\right)\pm √(\left(-2\right)^2-4\cdot \:1\cdot \:5))/(2)

Add 4 + -20:


x=(-\left(-2\right)\pm \:4i)/(2)

* the opposite of -2 is 2:


x=(2\pm 4i)/(2)

Now, solve equation:


x=(2+4i)/(2)

Divide 2 + 4i by 2:


x=1-2i

_____________________________

User Wandering Logic
by
8.3k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories