How can this polynomial have roots?

Algebra Level 3

Given the following expression:

$x^{2}-2x(m+2)+4m+13=0$ ( $m$ is a parameter)

Find $m$ so that we can find at least a real root $x$ of the given expression.

Choose the most correct answer.

Clarification: The answer "None are correct" refers to the answer A, B, C and D (in other words, rewrite "none are correct" as "A, B, C and D are incorrect".(or "This polynomial doesn't have any real roots")

m \ne \pm 3

m \leq -3

A,B,C and D are correct, but these happen individually D.

m \geq 3

None are correct. C and D are correct, but these happen individually A.

-3 \leq m \leq 3

A and B are correct, but these happen simultaneously

1 solution

Tom Engelsman
May 31, 2019

By the Quadratic Formula, the roots compute to:

$x = \frac{2(m+2) \pm \sqrt{4(m+2)^2 - 4(1)(4m+13)}}{2} = (m+2) \pm \sqrt{m^2 - 9}$

and in order in to obtain at least one real root, we require the discriminant to be non-negative $\Rightarrow m^2 - 9 \ge 0 \Rightarrow \boxed{m \in (-\infty, -3] \cup [3, \infty)}.$ Thus choices C and D are the most correct.

How can this polynomial have roots?

1 solution

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