Asked by Jos
How many real number solutions exist for 2x^2 + 8x + 8 = 0
The first part is 2x to the power of 2.
The first part is 2x to the power of 2.
Answers
Answered by
Reiny
2x^2 + 8x + 8 = 0
x^2 + 4x + 4 = 0
discriminant = b^2 - 4ac
= 16 - 4(1)(4)= 0
so there is 1 real answer
check:
x^2 + 4x + 4 = 0
(x+2)^2 = 0
x+2=0
x = -2
x^2 + 4x + 4 = 0
discriminant = b^2 - 4ac
= 16 - 4(1)(4)= 0
so there is 1 real answer
check:
x^2 + 4x + 4 = 0
(x+2)^2 = 0
x+2=0
x = -2
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