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2 Math questions
Quadratic Equations

2 Math questions Quadratic Equations-example-1
2 Math questions Quadratic Equations-example-1
2 Math questions Quadratic Equations-example-2
User OrPaz
by
7.9k points

2 Answers

7 votes

Answer:

It's the third Option.

Explanation:

It can't be the first 2 because they have imaginary roots not complex, because there is no term in x.

x^2 - 16x + 73 = 0

x^2 - 16x = -73

Completing the square:-

(x - 8)^2 - 64 = -73

(x - 8)^2 = - 73 + 64 = -9

x - 8 = +/- √-9 = +/- 3i

x = 8 +/- 3i.

User Arranjdavis
by
7.9k points
3 votes
The third one.

If we say that a quadratic eqtn takes the form of ax^2 + bx + c = 0, then in this eqtn a=1, b=-16 and c=73. Since the formula for finding the roots of quadratic eqtns is:

(-b+-sqrt(b^2-4ac))/2a, then:

(16 +- sqrt(16^2-4*73*1))/2

(16 +- sqrt(256-292))/2

(16 +- sqrt(-36))/2

Now, we know that the imaginary number “i” is equal to the square root of minus one. If we break down sqrt(-36) to sqrt(-1)*sqrt(36), then we get “6i” because the sqrt of 36 is 6 and the sqrt of -1 is i.

Now we have:

(16 +- 6i)/2 — if we divide this expression by two, we get 8 +- 3i.
User June Skeeter
by
8.6k points

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