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What is the solution set of the equation using the quadratic formula?

SamJones  Jan 26, 2018
 #1
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Solve for x:

x^2 + 6 x + 10 = 0

 

Subtract 10 from both sides:

x^2 + 6 x = -10

 

Add 9 to both sides:

x^2 + 6 x + 9 = -1

 

Write the left hand side as a square:

(x + 3)^2 = -1

 

Take the square root of both sides:

x + 3 = i or x + 3 = -i

 

Subtract 3 from both sides:

x = -3 + i or x + 3 = -i

 

Subtract 3 from both sides:

x = -3 + i      or      x = -3 - i

Guest Jan 26, 2018
 #2
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Solve for x:

x^2 + 6 x + 10 = 0

 

Using the quadratic formula, solve for x.

x = (-6 ± sqrt(6^2 - 4×10))/2 = (-6 ± sqrt(36 - 40))/2 = (-6 ± sqrt(-4))/2:

x = (-6 + sqrt(-4))/2 or x = (-6 - sqrt(-4))/2

 

Express sqrt(-4) in terms of i.

sqrt(-4) = sqrt(-1) sqrt(4) = i sqrt(4):

x = (-6 + i sqrt(4))/2 or x = (-6 - i sqrt(4))/2

 

Simplify radicals.

sqrt(4) = sqrt(2^2) = 2:

x = (-6 + i×2)/2 or x = (-6 - i×2)/2

 

Factor the greatest common divisor (gcd) of -6, 2 i and 2 from -6 + 2 i.

Factor 2 from -6 + 2 i giving -6 + 2 i:

x = (2 i - 6)/2 or x = (-6 - 2 i)/2

 

Cancel common terms in the numerator and denominator.

(2 i - 6)/2 = -3 + i:

x = i - 3 or x = (-6 - 2 i)/2

 

Factor the greatest common divisor (gcd) of -6, -2 i and 2 from -6 - 2 i.

Factor 2 from -6 - 2 i giving -6 - 2 i:

x = -3 + i or x = (-(2 i) - 6)/2

 

Cancel common terms in the numerator and denominator.

(-(2 i) - 6)/2 = -3 - i:

x = -3 + i        or        x = -i - 3

Guest Jan 26, 2018

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