Recall that (x-a)^2 = x^2-2ax+a^2
k^2-18k+50 = 0
k^2-18k = -50
Now, you want half of the coefficient of k and square it, so you have a perfect square. Then, to keep things equal, you have to add it to the right side as well.
k^2-18k+81 = -50+81
(k-9)^2 = 31
k-9 = ±√31
k = 9±√31
Now do the other the same way.
Solve by completing the square ( find the vertex only)
K^2-18k+50=0
M^2+500m+100=0
2 answers
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