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stira [4]
3 years ago
5

CAN SOMEBODY PLEASE ANDWER THIS!! Given the function f(x)= 14x^2 + 5x - 1 if you could change the 14 to any other number, but th

e rest has to stay the same, what value would it have to be so that there is only 1 real root for the function? Show or explain.
Mathematics
1 answer:
dybincka [34]3 years ago
4 0

a {x}^{2}  + 5x - 1 = 0 \\ 25 + 4a = 0 \\ a =  \frac{ - 25}{4}  =  - 6.25
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Answer:

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Step-by-step explanation:

The equation of a circle in standard form is

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Given area = 25π , that is

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r² = 25 ( take the square root of both sides )

r = \sqrt{25} = 5

With centre (h, k) = (3, - 9 ) and r = 5 , then

(x - 3)² + (y - (- 9) )² = 5² , that is

(x - 3)² + (y + 9)² = 25

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