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Stuck solving this equation.

Full text:

For what real values of the parameter do the common solutions of the equation became identical?

1. y = mx - 1
2. x^2 = 4y
ans. m = +/- 1

2a. x^2 - 4y = 0

So i started by substituting 1. into 2a.

x^2 - 4 * (mx - 1) = 0
x^2 -4mx + 4 = 0

Solutions must be identical so we must have x1 = x2 and y1 = y2

From the book: Therefore the two common solutions of (1) and (2) become identical when and only when the roots of the equation are equal; that is when the discriminant vanishes

d = B^2 - 4AC

A = x^2
B = -4mx
C = 4

Here i got stuck

d = (-4mx)^2 - 4 * x^2 * 4
(-4mx)^2 - 4 * x^2 * 4 = 0
(-4mx)^2 + 16 * x^2 = 0

looking at previous solutions, i should get the value of m from determinant by getting a trinomial that i could factor. I am unable to form a trinomial. So how can this be solved?

Martin Berger
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3 Answers3

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The discriminant of a quadratic in $\,x\, $ does not involve $\,x\,$ but only the coefficients, so

$$x^2-4mx+4=0\implies \Delta=(-4m)^2-4\cdot1\cdot4=16m^2-16=0\iff$$

$$\iff 16m^2=16\iff m^2=1\iff m=\pm\,1$$

DonAntonio
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    And i was wondering why i was unable to solve last 5 or so equations... I was including x in A, B and C. I completely forgot about that. Thank you very much for helping :) – Martin Berger Jun 01 '13 at 14:38
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The values of A, B and C are the coefficients of $x^2, x$ and the constant term respectively but do not include the '$x$' parts. This should solve your problem.

john
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It is not necessary to use the discriminant here, since it is easy to complete the square (an equivalent procedure).

$$0=x^2-4mx+4=(x-2m)^2+4-4m^2$$ and this has equal roots when $4-4m^2=0$. Note, this is one quarter of the discriminant.

Mark Bennet
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