Solve for k, given Y

_{Parabola}= 1x^{2}+ (2k + 10)x + k^{2}+ 5, tangent to the x-axis. Mary says “Obviously”, b

^{2}– 4ac = 0 (Johnny says, "This may not be obvious. I will provide a note below.) (2k + 10) ^{2} – 4(1)( k^{2} + 5) = 0 | Substitution of a = 1, b = 2k + 10 and c = k ^{2} + 5 |

4k ^{2} + 40k + 100 – 4k^{2 } – 20 = 0 | Expand |

40k = -80 | |

k = -2 | Done. |

Note: At the point where a line is tangent to a parabola, Y

_{Parabola}= Y_{Line}._{ }Therefore Y_{Parabola }– Y_{Line}= 0_{. }But in this case the tangent line is the x-axis (given) so the line is simply y = 0. (or, y = 0x + 0 if you want to be precise!)_{ }Since the gradient of y = 0 must be m = 0, we know the tangent is horizontal. This only happens at a stationary point. But a parabola only has one stationary point (the maximum or minimum). In this case the stationary point touches the x-axis so it is also the only root of the parabola. Consequently the discriminant, b^{2}– 4ac is 0. (Recall that, if there are two real roots b^{2}– 4ac > 0, if there are no real roots b^{2}– 4ac < 0.)Calculus has only been used

*in the background*to understand the problem. We never found the first derivative. This suggests an alternative but more complicated approach.Alternate method:

m = dy/dx = 2x + 2k + 10

As noted above, m = 0 (The tangent is horizontal).

0 = 2x + 2k = 10

0 = 2(x + k + 5)

0/2 = 0 = x + k + 5

x = – k – 5

Substitute this x into Y

_{Parabola }= 0 (The tangent is the x-axis which is the line y = 0).1x

^{2}+ (2k + 10)x + k^{2}+ 5 = 0(–k – 5)

^{2}+ (2k + 10) (–k – 5) + k^{2}+ 5 = 0k

^{2}+ 10k + 25 – 2k^{2}– 10k –10k – 50 + k^{2}+ 5 = 020 = -10k

k = -2 As before. Undoubtedly there exist even more complicated solutions but Johnny will stop here.

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