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For two real values of n, the equation 4x^2+nx+25 has exactly one solution in x. What is the positive value of n?

Guest Jun 16, 2018
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there are only a few ways to setup this into factored form

 

 

(2x+-25)(2x+-1)

(4x+-1)(x+-25)

(4x+-25)(x+-1)

(4x+-5)(x+-5)

(2x+-5)(2x+-5)

 

if we want sqrt of B^2-4ac to have one solution that has to equal 0

 

the only one where B^2 = 400 is when b = 20 or (2x+-5)(2x+-5)

 

so N or b = 20

Guest Jun 16, 2018

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