What reduced quadratic equation has solutions 2 and 3?

William Collins

William Collins

Answered

2022-01-13

What reduced quadratic equation has solutions 2 and 3?

Answer & Explanation

aquariump9

aquariump9

Expert

2022-01-14Added 40 answers

Step 1
A polynomial with coefficients in Q that has 2 as a root must also have 2 as a root; likewise, if it has 3 as a root, then it must also have 3 as a root.
So any polynomial with coefficients in Q that has both 2 and 3 as roots must be of degree at least 4 or equal to 0, and must be a multiple of (x22)(x23).
So there is no quadratic polynomial with coefficients in Q that will do it.
Step 2
To see this, suppose that p(x) is a polynomial with coefficients in Q that has 2 as a root. Dividing with remainder by x22 we get
p(x)=(x22)q(x)+r(x),
where q(x) and r(x) have rational coefficients, and r(x)=0 or else r(x) is of degree 0 or 1.
Plugging in 2, we get
0=0q(2)+r(2)=r(2)
But there are no polynomials of degree 0 or 1 with coefficients in Q that are 0 at 2 (that would yield that 2 is rational), so r(x)=0.
Therefore, r(x)=0, so p(x) is a multiple of x22; and since x22 is 0 at 2, then so is p(x). The same argument shows that it is also a multiple of x23, and so p(x) will also have 3 a a root.
Step 3
Since x23 and x22 are relatively ' in Q[x], any polynomial that is a multiple of both is a multiple of their product. The smallest degree polynomial that doe is
(x22)(x23)=x45x2+6
If you allow other coefficients, then the only quadratics that work are the form
c(x2)(x3)
=cx2x(2+3)x+6
and its multiples.

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