Tricky Quadratics

Algebra Level 5

Let $m$ be a constant real number . Find the sum of all the real roots of the equation $x^2 - 2mx + (m^2 + 1) = 0$ .

2m

0

-2m

None of the other choices

3 solutions

Tom Engelsman
Dec 29, 2016

The above quadratic equation factors into $[x - (m+i)][x - (m-i)] = 0$ with the complex-conjugate pair of roots $x = m \pm i.$ Thus there are NO real roots to sum together.

As I understand, the sum of an empty sum of numbers is, by convention, $0$ .

Brian Charlesworth - 4 years, 5 months ago

Yashas Ravi
Feb 12, 2019

We can group $(x^2 - 2mx + m^2) + 1= 0$ into $(x-m)^2 + 1$ . Without adding $1$ to $(x-m)^2$ , $(x-m)^2$ is a quadratic that touches the $x$ -axis at $1$ point, which is the vertex. By adding $1$ , we increase the $y$ -coordinate by $1$ , so the vertex no longer touches the $x$ -axis. Because there are now no $x$ -intercepts, the equation has $0$ solutions.

Edwin Gray
Aug 23, 2018

The discriminant = -4, so there are 0 real solutions. Ed Gray

Tricky Quadratics

3 solutions

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