Quadratic Function
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Dr.Peterson
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If you are asking about #4, you might start by supposing that the roots are r and 3r, and writing the equation in factored form.
Otis
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Not in this exercise. 3r = r when y=x^2, but we can't express a in terms of b and c, for that polynomial.
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Otis
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The quadratic formula gives the two roots, in terms of parameters a,b,c. You could multiply one of them by 3 and set the result equal to the other. Solve that equation for a, for b and for c.
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Steven G
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Why not y = ax^2 + bx?? Recall if 3r = r then r= 0 and x= r =0 is a root of y=ax^2 + bx.
It is a silly point because in that case the quadratic will have '3' roots. x=0, x= 3*0 and x= -b/a
Otis
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It seems that you answered your own question, Jomo. The roots of ax^2+bx are 0 and -b/a.
Case 1: 3(0) = -b/a
Case 2: 3(-b/a) = 0
In either case, b must be zero. So we're left with ax^2. (I ought to have written y = ax^2, instead.)
Of course, in the given exercise we have 3·r1 = r2
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