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22ksotoq
03.12.2020 •
Mathematics
What value(s) of b will cause 3x2 + bx + 3 = 0 to have one real solution? Select all that apply.
b = –6
B.b = –9
C.b = 6
D.b = 9
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Ответ:
-6 and 6
Step-by-step explanation:
Possible factors that could produce 3x² + 6x +3 are (x + 1)(3x + 3)
the solution for values of x: x+1=0, x= -1 3x+3=0, 3x= -3 . x= -1 . Same.
Possible factors that could produce 3x² - 6x +3 are (x - 1)(3x - 3)
the solution for values of x: x-1=0, x= 1 3x-3=0, 3x= 3 . x= 1 . Same.
From the attached graph of the possible equations, you can see that 9 and -9 won't produce ONE real solution. They each have "zeros" at two points on the x-axis. That is two solutions that would make x = 0.
The equations with -6 and +6 each touch the x-axis at only one point each.
Ответ:
Step-by-step explanation:
The third side is 6.
Hope it helps