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3 years ago

What are the solutions of x^2-2x+10=0

2 answers:

8 0

There are a few ways to solve this, the simplest of which is probably factoring. Start by finding factors of C (10) that add up to be B (-2).

fac(10) { -10, 1 ; -5, 2 }

Well, there are no factors of 10 that can add to b - so we will use the quadratic formula.

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

where a = 1, b = -2, c = 10

b^2 - 4ac = (-2)^2 - 4(1)(10) = 4 - 40 = -36

-b +- sqrt(-36) / 2a

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

sqrt(-36)

sqrt(-1 * 36)

sqrt(-1) * sqrt(36)

i * 6

2 + 6i / 2 ; 1 + 3i

2 - 6i / 2 ; 1 - 3i

Solutions: 1 + 3i, 1 - 3i

cluponka [151]3 years ago

7 0

Answer:

x = 1± 3i

Step-by-step explanation:

x^2-2x+10=0

We can complete the square to solve by subtracting 10 from each side

x^2-2x+10-10=-10

x^2 -2x = -10

We need to add (2/2) ^2 to each side or 1

x^2 -2x+1 = -10 +1

x^2 -2x+1 = -9

The left side factors into (x- (2/2) ) ^2

(x-1) ^2 = -9

Take the square root of each side

sqrt((x-1) ^2 =± sqrt(-9)

x-1 = ±sqrt(-1) sqrt(3)

Remember the sqrt(-1) = i

x-1 = ± 3i

Add 1 to each side

x-1+1 = 1± 3i

x = 1± 3i

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4 years ago

Rectangle ABCD has vertices at (-7,-2); (1, -2); (1, -8); and (-7, -8) respectively. If GHJK is a similar rectangle where G(2, 5

Gennadij [26K]

Answer:

Points J and K could be located at:

J(2,2)

K(6,2)

Step-by-step explanation:

Consider the vertices have a x and y coordinate:

A: x coordinate=-7 y coordinate=-2

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