x=(-b±√(b^2-4ac))/2a

The cry is of course, “When will we ever use this?” sometimes restated as, “We aint never ever gonna need this!”

Maths is not real. It’s all imaginary. If you are a realist, you will never use the quadratic formula. But if you do use the quadratic formula, chances are you will use it in situations like this one.

*A circle of radius 10 exists. Each of four congruent smaller circles are tangent to the outer circle and to same-sized neighbours. The smallest circle has the same centre as the large circle and is tangent to all four intermediate-sized circles. What is the radius of the smallest circle? See diagram. (Hint: you might want to use the quadratic formula.)*

A circle is not a parabola, so we appear to be a long way from the quadratic formula. However, the formula of both is second degree. A ‘plain vanilla’ circle is x

^{2}+ y

^{2}= r

^{2}. Likewise, the most basic parabola is y = x

^{2}. The teacher writes both of these formulas on the board.

Johnny and Mary discuss the problem. Mary notes the four intermediate circles create 4-fold symmetry. Johnny wants to use Cartesian coordinates. Together they realise the lines of symmetry must be the x-axis, the y-axis, the line y = x and the line y = -x. The teacher agrees and suggests that the use the symmetry to reduce the problem to the first quadrant. He also asks, “What is the distance from O to A?” They redraw the problem as folows.

Johnny says the formula of the circle reminds him of Pythagoras: a

^{2}+ b

^{2}= c

^{2}. Mary suggests the distance formula, D = √ [(x2 – x1)

^{2}+ (y2 – y1)

^{2}] is likely to be important. Another student in the group says he prefers to write D2 = (x2 – x1)

^{2}+ (y2 – y1)

^{2}, thus avoiding the square root sign.

The origin is (0, 0). Point A is on the line y =x, so A = (xA, yA) = (xA, xA). The distance from the origin to point A is 10, so D

^{2}= 100. Thus (x2 – x1)

^{2}+ (y2 – y1)

^{2}becomes (xA – 0

^{2}+ (y2 –0)

^{2}= (xA – 0)

^{2}+ (xA –0)

^{2}. Therefore 100 = 2xA

^{2}.

xA = √50 = 5√2.

What are the coordinates of B?

Just like point A, B is also on the line y = x. Consequently, B = (xB, yB) = (xB, xB) = (x, x) for ease of writing. The square of the distancefrom A to B is DAB

^{2}= (5√2 – x)

^{2}+ (5√2 – x)

^{2}. The square of the distancefrom BxB to B is D2 = (x – x)

^{2}+ (0 – x)

^{2}. (Not a misprint, remember y = x). Since AB and BxB (or ByB) are both radii, they are the same distance.

Therefore, D

^{2}= DAB

^{2}

(x– x)

^{2}+ (0 – x)

^{2}= (5√2 – x)

^{2}+ (5√2 – x)

^{2}

0

^{2}+ x

^{2}= (50 – 10√2x + x

^{2}) + (50 – 10√2x + x

^{2})

x

^{2}= 100 – 20√2x + 2x

^{2}

0 = 100 – 20√2x + x

^{2}

At last we are ready to apply the quadratic formula, with a = 1, b =–20√2 and c = 100. Then the required radius, OA

**’**= 10 – 2x = 1.71572875…(Google calculator)

Whereas, a realist would have made a scale drawing and found OA’ = 1.7 to 2 significant figures.

Who uses the quadratic formula? Dreamers and idealists. Don’t be too dismissive George Bernard Shaw wrote:

The reasonable man adapts himself to the world; the unreasonable one persists in trying to adapt the world to himself. Therefore all progress depends on the unreasonable man.

Of course there are a few loose ends. It is justified to reduce the problem to a single quadrant? Further, is it justified to insist that y = x is the axis of symmetry for the reduced problem?But these are thoughts for another day.

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