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What are the roots of this equation?

x²-4x+9=0

a. 1±i√5

b. 2±i√5

c. 2±i√3

d. 1±i√52​

User Rxin
by
4.9k points

1 Answer

3 votes

Answer:

x = 2 ± (i)√5 (Answer b)

Explanation:

x²-4x+9=0 can be solved in a variety of ways; the first two that come to mind that are also appropriate are (1) completing the square and (2) using the quadratic formula.

Completing the square is fast here:

Rewrite x²-4x+9=0 as x²-4x +9=0

Identify the coefficient of the x term: it is -4

Take half of that, obtaining -2

Square this result, obtaining 4

Add 4 to x²-4x +9=0, in the blank space in the middle, and then subtract 4: x²-4x +4 -4 +9=0

Rewrite x²-4x +4 as the square of a binomial:

(x - 2)² - 4 + 9 = 0 → (x - 2)² = -5

Take the square root of both sides: x - 2 = ±√(-5) = ± (i)√5

Then x = 2 ± (i)√5

User Fabian Knorr
by
5.5k points
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