Find k such that the line y = 5x + 3 is tangent to the graph of the function below. y=ksqrt(x)?

Question

what does k=?

Answer 1

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y = k√x

y' = k / 2√x

Now if the given line is tangent to y = k√x

then the slope of the line equals k / 2√x

so k / 2√x = 5

k = 10√x [eqn 1]

Now to find the point where tangent touches the curve:

k√x = 5x + 3 [eqn 2]

Subs [eqn 1] into [eqn 2] → 10√x . √x = 5x + 3

10x = 5x + 3

5x = 3

x = 3/5

Now subs x = 3/5 into [eqn 1] → k = 10√(3/5)