Solution to #10 on Winter Break Review Packet
If 3x^2+20x+k=0 has exactly one solution, then find the value of k.
The way we approach this problem is to set the discriminant equal to 0, and then solve for k.
sqrt(b^2-4*a*c)=0
sqrt(20^20-4*3*k)=0
20^20-4*3*k=0
-12k=-400
k=100/3
The way we approach this problem is to set the discriminant equal to 0, and then solve for k.
sqrt(b^2-4*a*c)=0
sqrt(20^20-4*3*k)=0
20^20-4*3*k=0
-12k=-400
k=100/3
posted by Anonymous at 2:54 PM
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