5x^2 - kx + 10
to have one solution, the discriminant must be zero
k^2 -200 = 0
now finish it off
For what value of k does kx − 10 = 5x^2 have one solution?
2 answers
5x^2 - kx + 10 = 0
To have one (actually 2 equal) solution, b^2 - 4ac = 0
(-k)^2 - 4(5)(10) = 0
k^2 = 200
<b<k = ± √200 = ± 10√2
check: using +√2
5x^2 - 10√2 x + 10 = 0
x^2 - 2√2 x + 2 = 0
(x - √2)^2 = 0
x - √2 = 0
x = √2 <------ one solution
checking for k = -10√2 will yield the ONE solution of x = - √2
To have one (actually 2 equal) solution, b^2 - 4ac = 0
(-k)^2 - 4(5)(10) = 0
k^2 = 200
<b<k = ± √200 = ± 10√2
check: using +√2
5x^2 - 10√2 x + 10 = 0
x^2 - 2√2 x + 2 = 0
(x - √2)^2 = 0
x - √2 = 0
x = √2 <------ one solution
checking for k = -10√2 will yield the ONE solution of x = - √2
9 answers