I think the correct answer is that it is impossible to write a real polynom having the prescribed characteristics. In fact, in order for $2$ and $-i$ to be roots and for the polynom to have real coefficients, also the conjugate of $x-i$ needs to be a root, so that: \begin{equation} P(x)=(x-2)(x-i)(x+i)Q(x) \end{equation} So the miniumum degree of $P$ is $3$. Or am I wrong?
Determine a polynom $P(x)\in\mathbb{R}[X]$ of degree $2$ such that $P(2)=0$ and $P(-i)=3$.
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$\begingroup$
linear-algebra
polynomials
complex-numbers
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0$P(-i)=3$ so $-i$ is *not* a root. – 2017-01-09
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0Note that $i$ isn't a root – 2017-01-09
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0set $$f(x)=ax^2+bx+c$$ – 2017-01-09
4 Answers
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You are wrong in saying that there is no such polynomial, note that $i$ is no root of $P$.
Let $P \in \mathbf R[X]$ with $P(-i) = 3$, then $-i$ is a root of $Q := P - 3$, therefore $i$ is also a root of $Q$, hence $Q$ has to be a multiple of $x^2 + 1$. As $Q$ is of degree 2, $Q(x) = a(x^2 + 1)$ for $a \in \mathbf R$. This gives $$ P(x) = a(x^2 + 1) + 3 $$ With $$0= P(2) = 5a + 3 \iff a = -\frac 35$$ we get $$ P(x) = -\frac 35(x^2 + 1) + 3. $$
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No, the minimum degree is not $3$, and there is indeed such a real polynomial. We may take $$ P(x)=-\frac{3}{5}(x-2)(x+2), $$ which satisfies $P(2)=0$ and $P(-i)=3$, because $-\frac{3}{5}(-i-2)(-i+2)=3$.
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$P(x) = a(x^2 + 1) + 3\\ P(2) = 5a + 3=0$
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If $P\in \mathbb{R}[X]$ has degree $2$, and $P(2)=0, P(-i)=3$, then $$ P(X)=a(X-2)(X-b)=a[X^2-(b+2)X+2b], $$ where $a$ and $b$ are such that $a, a(b+2), ab \in \mathbb{R}$, with $a\ne 0$. Since $$ 3=P(-i)=a[-1+(b+2)i+2b]=-a+2ab+a(b+2)i $$ we have $$ -a+2ab=3,\quad a(b+2)=0 $$ and deduce that $b=-2$ because $a\ne 0$, and therefore $a=-3/5$.
Thus $$ P(X)=-\dfrac35(X-2)(X+2)=\dfrac{12}{5}-\dfrac{3}{5}X^2. $$