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Mathematics 19 Online
OpenStudy (dtan5457):

4x^2+kx+25=0 Find Values of k that ensure that the given equation has exactly one solution

OpenStudy (dtan5457):

Anyone get this?

jimthompson5910 (jim_thompson5910):

4x^2+kx+25=0 is in the form ax^2+bx+c=0 where a = 4, b = k, c = 25

jimthompson5910 (jim_thompson5910):

"Find Values of k that ensure that the given equation has exactly one solution" so this means the discriminant D is equal to 0 The discriminant formula is D = b^2 - 4ac

jimthompson5910 (jim_thompson5910):

plug in a = 4, b = k, c = 25 and D = 0 then solve for k

OpenStudy (dtan5457):

K+/- 20

OpenStudy (dtan5457):

so if there wasn't one solution? you can use quadratic formula?

jimthompson5910 (jim_thompson5910):

yes k = 20 or k = -20 will make 4x^2+kx+25=0 have exactly one solution for x

jimthompson5910 (jim_thompson5910):

if you want 4x^2+kx+25=0 to have exactly 2 solutions, then you would look for values of k that make D > 0 true if you want 4x^2+kx+25=0 to have no real solutions, then you would look for values of k that make D < 0 true

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