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anonymous
 4 years ago
Find the quadratic equation whose solutions are:
2 + sqr(5) and 2  sqr(5)
I'm not just looking for an answer but rather to learn.
anonymous
 4 years ago
Find the quadratic equation whose solutions are: 2 + sqr(5) and 2  sqr(5) I'm not just looking for an answer but rather to learn.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[2+ \sqrt{5} and 2  \sqrt{5}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0equation bcam x^2  (a+b)x+ a.b =0 take a=2+sqr 5 , b=2sqr 5

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[x ^{2}  (a+b)x + ab = 0?\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0this is from an example in my book, but I don't understand the example:

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0tis what any quarratic equation can be form...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if their two roots are given .got it ...now

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1326657719757:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that's the example from the book

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yea ........now plug in...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If you are given 2 (or more than 2, it doesn't matter) roots of a polynomial, say a and b in this case, then that just means if you were to factor your polynomial, you would get \[f(x)=(xa)(xb)\] By multiplying this together, you get the equation \[x^2(a+b)+ab\] Thus, in your example, you would just plug in \[a=2+\sqrt{5}, b=2\sqrt{5}\] to the above equation to reach your answer.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1326657923682:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that's the books example typed out,

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thank you, i'm trying it
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