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The polynomial \begin{equation*} p(x)= ax^2+bx+c \end{equation*} has $1+\sqrt{3}$ as one of it's roots and also $p(2)=-2$. Is there any way to know the value of $a$, $b$ and $c$? I tried but I can form only $2$ equations how can I figure out value of $3$ unknown with $2$ equations so something must be missing.

Also $a$, $b$ and $c$ are rationals.

bulbasaur
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3 Answers3

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Hint: If one of the roots is $1 + \sqrt{3}$ and $a, b, c$ are rational then what must the other root be?

Solution 1:

The other root must then be $1 - \sqrt{3}$.

You have two options now: You can make a system of three linear equations in three variables and simply solve.

Or, you can simply expand $\alpha[x - (1 - \sqrt{3})][x - (1 + \sqrt{3})] = \alpha(x^2 - 2x - 2)$ and use $p(2) = -2$ to get $\alpha = 1$.

Solution 2:

We only need two equations if we know that $a, b, c$ are rational.

$$(1+\sqrt{3})^2 a + (1+\sqrt{3})b + c = 0$$

$$4a + 2\sqrt{3} a + b + \sqrt{3} b + c = 0$$

$a, b, c$ are rational, so we know that $-2a = b$ because the $\sqrt{3}$ terms must cancel each other. So, $4a + 2b + c = -2$ gives us $c = -2$.

Substituting back into the above we get $4a + b = 2$, so $a = 1$ and $b = -2$.

Thus the quadratic is $x^2 - 2x - 2$.

MT_
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  • +1 for the hint. You could say $p(x)=\alpha (x-x_1)(x-x_2)=\alpha (x^2-2x-2)$ and use $p(2)=-2$ to conclude $\alpha=1$. – mhp May 22 '15 at 07:49
  • @mhp Right, forgot about the $p(2)= -2$ condition to get the lead coefficient – MT_ May 22 '15 at 14:39
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$$p(x)=ax^2+bx+c$$ given that $x=1+\sqrt 3$ so there must me an $x=1- \sqrt 3$ as in a quadratic polynomial when you get the roots the irrational part is from the discriminant as $a,b,c$ are rational. $$p(x)=(x-(1+\sqrt 3))(x-(1- \sqrt 3))$$ expanding this we get, $$p(x)=x^2-2x-2$$

Ilaya Raja S
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  • I believe you mean discriminant rather than determinant, also you said $1+\sqrt{3}$ twice. Perhaps you meant $1-\sqrt{3}$ the second time? – jgon May 22 '15 at 04:39
  • ya discrimi.. that thing and thanks for pointing the type error – Ilaya Raja S May 22 '15 at 04:40
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Let $x = 1+\sqrt{3} \Rightarrow (x-1)^2 = 3 \Rightarrow x^2-2x-2 = 0 \Rightarrow a = k, b = -2k,c = -2k$, and $p(2) = -2 \Rightarrow 4a+2b+c=-2\Rightarrow 4k - 4k - 2k = -2 \Rightarrow k = 1 \Rightarrow (a,b,c) = (1,-2,-2)$.

DeepSea
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  • Does this implicitly use the fact that $a, b, c$ are rational anywhere? – MT_ May 22 '15 at 04:27
  • I am struggling to see how only the fact that $p(2) = -2$ and $1 + \sqrt{3}$ is a root, with all coefficients in the reals, gives us a unique quadratic. Why can't we construct many such quadratics with that property? – MT_ May 22 '15 at 04:32
  • We can, we only have two points. For any three points (with distinct $x$ values), there is a unique quadratic through those points by Lagrange interpolation. – jgon May 22 '15 at 04:33
  • @jgon Which is why ask how NotALoner gets a unique solution with only two points – MT_ May 22 '15 at 04:35
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    This answer is incomplete. If you do not use the fact that the coefficients are rational, you could equally well say$$x=1+\sqrt3\quad\Rightarrow\quad(x-1-\sqrt3)(x+1)=0\quad\Rightarrow\quad x^2-\sqrt3x-1-\sqrt3=0$$and so$$(a,b,c)=k(1,-\sqrt3,-1-\sqrt3)\ .$$ – David May 22 '15 at 04:35
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    You're tacitly assuming that $x^2-2x-2$ is the minimal polynomial for $1+\sqrt{3}$, so that it divides $ax^2+bx+c$, so that $a=k$, $b=-2k$, and $c=-2k$. This is true over $\Bbb{Q}$ not $\Bbb{R}$. – jgon May 22 '15 at 04:36
  • @David I thought something fishy was at play here. – MT_ May 22 '15 at 04:36
  • @Soke Your hint is correct, I have upvoted it. – David May 22 '15 at 04:37