Finding roots

Steven G

Steven G

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Suppose you are given the equation 6x^2+5x+1 = 0. How would you solve this?
 
Steven G said:
Did you notice that you got the reciprocal of -2 and -3? Do you know why?
Click to expand...
Well, I got the answer from solve the 2 equations
  1. (3x+1)=0 =>3x=-1 =>x=-1/3
  2. (2x+1)=0 =>2x=-1 =>x=-1/2
I think this would happen when the "c" in the quadratic is 1, and the quadratic is factorable.
 
AvgStudent said:
Well, I got the answer from solve the 2 equations
  1. (3x+1)=0 =>3x=-1 =>x=-1/3
  2. (2x+1)=0 =>2x=-1 =>x=-1/2
I think this would happen when the "c" in the quadratic is 1, and the quadratic is factorable.
Click to expand...
Can you think of a quadratic that has x=-3 and x=-2 as its roots?
 
Steven G said:
Does the quadratic you just got and the one which I started this post with look similar in any way?
Click to expand...
Yes, very interesting. It's a different combination of the coefficients 1, 5, and 6, more specifically "a" and "c" are swapped while "b" remained the same. Is that the point you're trying to get at?
Suppose we have a quadratic [imath]ax^2+bx+c=0[/imath], and its solutions are [imath]x_1[/imath] and [imath]x_2[/imath],
then the roots of [imath]cx^2+bx+a=0[/imath] are [imath]\frac{1}{x_1}[/imath] and [imath]\frac{1}{x_2}[/imath]
 
AvgStudent said:
Yes, very interesting. It's a different combination of the coefficients 1, 5, and 6, more specifically "a" and "c" are swapped while "b" remained the same. Is that the point you're trying to get at?
Suppose we have a quadratic [imath]ax^2+bx+c=0[/imath], and its solutions are [imath]x_1[/imath] and [imath]x_2[/imath],
then the roots of [imath]cx^2+bx+a=0[/imath] are [imath]\frac{1}{x_1}[/imath] and [imath]\frac{1}{x_2}[/imath]
Click to expand...
You got it. Now try to prove it! Of course \(\displaystyle c \neq 0\)
 
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