SOLUTION: Find the qundratic equation whose roots are (3+sqrt(5)) and (3+squt(5))

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Question 717790: Find the qundratic equation whose roots are (3+sqrt(5)) and (3+squt(5))
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
I suspect "qundratic" should be "quadratic" and "(3+sqrt(5)) and (3+squt(5))" should be "(3+sqrt(5)) and (3-sqrt(5))". Please be careful when posting. Tutors are less likely to reply if it seems you didn't care enough to bother to type in the problem accurately.

If a number, let's call it "r", is a root of a polynomial, then (x - r) is a factor of that polynomial. So if %283%2Bsqrt%285%29%29 and %283-sqrt%285%29%29 are roots of the desired equation then %28x+-%283%2Bsqrt%285%29%29%29 and %28x-%283-sqrt%285%29%29%29 are factors of it. We can use this to write out equation in factored form:
y+=+%28x+-%283%2Bsqrt%285%29%29%29%2A%28x-%283-sqrt%285%29%29%29
Simplifying the factors:
y+=+%28x+-3-sqrt%285%29%29%29%2A%28x-3%2Bsqrt%285%29%29%29
Next we multiply this out. The methodical way is to multiply each term of one factor times each term of the other and then multiplying it out. But if we use some factoring patterns we can do this much more quickly and easily.

One of the factoring patterns you should know is:
%28a%2Bb%29%28a-b%29+=+a%5E2-b%5E2
Although this is usually used to factor (i.e. "un-multiply") expressions, it can also be used in the other direction to multiply. With a little imagination we should be able to see that the two factors fit the (a+b)(a-b) pattern, with the "a" being "x-3" and the "b" being "sqrt%285%29". (Take a moment to look at the factors and see this pattern.) So according to the pattern we should get a%5E2-b%5E2:
y+=+%28x+-3%29%5E2-%28sqrt%285%29%29%5E2
The square root is simple to square:
y+=+%28x+-3%29%5E2-5
Squaring the (x-3) is not so simple. But for this we can use another pattern, %28a-b%29%5E2+=+a%5E2-2ab%2Bb%5E2, to multiply %28x-3%29%5E2. With the "a" being "x" and the "b" being "3" we get:
y+=+x%5E2+-2%28x%29%283%29+%2B%283%29%5E2-5
which simplifies as follows:
y+=+x%5E2+-2%28x%29%283%29+%2B9-5
y+=+x%5E2+-6x+%2B9-5
y+=+x%5E2+-6x+%2B4
This is a quadratic equation with the specified roots.

P.S. The equation we found is not the only possible correct answer. These equations, in addition to the (x-r) factors, can have a non-zero constant factor. For example we could have used:
y+=+2%2A%28x+-%283%2Bsqrt%285%29%29%29%2A%28x-%283-sqrt%285%29%29%29
If so we would end up with:
y+=+2x%5E2+-12x+%2B8 (which also has the specified roots)
Or
y+=+-7%2A%28x+-%283%2Bsqrt%285%29%29%29%2A%28x-%283-sqrt%285%29%29%29
ending up with:
y+=+-7x%5E2+%2B42x+-28 (which also has the specified roots)
etc.