question (1 Viewer)

[bos1234]

Show that if the qun ax^2+bx+c =0 has

(b) the roots equal in absolute valuye but opposite in sign, then b=0

(c) the roots are reciprocals of one another, when c = a

(d) the roots with opposite signgs, then c<0 when a > 0

thanks bye

 

[conman]

Let x1 is one of the roots of the equation
x2 is another root of the equation

Now, we know that x1 + x2 = -b/a
x1x2= c/a
Use this to solve all the a, b, c questions. Hope u will come up with an answer!!!

 

[AlvinCY]

Another way of looking at it is this:


a) let the roots be A and – A , the sum of roots would prove b = 0
b) let the roots be A and 1/A, the product of roots would give you c = a
c) let the roots be A and - A, then the product of roots is -A² = c/a, since -A² is always negative, then when a > 0, c must be < 0


QED.

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[airie]

c) let the roots be A and - A, then the product of roots is -A² = c/a, since -A² is always negative, then when a > 0, c must be < 0[FONT=&quot][/FONT]

But Alvin, you're assuming that the roots have equal absolute value here. Just let the roots be A and -B, where A and B are positive, then their product -AB must be negative. Thus is a>0, c<0.

 

[AlvinCY]

Haha... pardon me. Yes, you're quite right, but regardless, A and -B would yield the same solution. (Where A and B are both positive numbers... or both negative... that way, A and -B are always opposite in sign, and doesn't matter whether A is negative or B is)

So yes, -AB, same argument.

Thanks Airie.