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How and why do i get 2,5 from square root of (1/2)^-6

 Sep 16, 2016
 #1
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Solve for x:
-x^2-x+6 = 0

The left hand side factors into a product with three terms:
-(x-2) (x+3) = 0

Multiply both sides by -1:
(x-2) (x+3) = 0

Split into two equations:
x-2 = 0 or x+3 = 0

Add 2 to both sides:
x = 2 or x+3 = 0

Subtract 3 from both sides:
Answer: |x = 2                or                x = -3

 Sep 16, 2016
 #2
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Use quadratic formula:
Solve for x:
-x^2-x+6 = 0

Multiply both sides by -1:
x^2+x-6 = 0

x = (-1±sqrt(1^2-4 (-6)))/2 = (-1±sqrt(1+24))/2 = (-1±sqrt(25))/2 = (-1±5)/2:
x = (-1+5)/2 or x = (-1-5)/2

(-1+5)/2 = 4/2 = 2:
x = 2 or x = (-1-5)/2

(-1-5)/2 = -6/2 = -3:
Answer: |x = 2 or x = -3

 

P. S. COMPARE YOUR SOLUTION TO THIS ONE USING QUADRATIC FORMULA.

 Sep 16, 2016

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