Question

Use the discriminant to determine the nature of the roots of the following equation.

x2 + 2x + 5 = 0

Double root
real and rational root
real and irrational root
imaginary root

Answer 1

A Quadratic Equation\[ax^2+bx+c=0\]
The Quadratic Formula\[x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
Factored for of the Quadratic Equation\[(x-x_1)(x-x_2)=0\]

start by identifying the coefficients a,b,c

Compare
\[x^2+2x+5=0\]
With
\[ax^2+bx+c=0\]

\(a=?\) \(b=?\) \(c=?\)


The radicand of the quadratic formula is called
the Discriminant \(\Delta =b^2-4ac\)

if \(\Delta=0\) there is only one solution to the quadratic equation
if \(\Delta>0\) there are two solutions
if \(\Delta <0\) the solutions are complex

another way of putting this
if \(\Delta=0\) there is two equal solutions
if \(\Delta>0\) there are two different solutions
if \(\Delta<0\) there are no real solutions

if \[\Delta \neq 0= d^2 \] the solutions are rational

then taking the square root in the quadratic formula will not leave any irrational component

\[x_{1,2}=\frac{-b\pm \sqrt\Delta}{2a}=\frac{-b\pm \sqrt d^2}{2a}=\frac{-b\pm d}{2a}\]

Answer 2

"No real roots" isn't one of the answer choices.

Answer 3

What are your values for \(a\), \(b\), \(c\)?

What do you get for the discriminant ?

\[\Delta=b^2-4ac\]

Answer 4

Okay, thanks!

Answer 5

What do you get?

Answer 6

\[\Delta = (2)^2-4\times(1)\times(5)=\dots\]

Answer 7

"No real roots" is the same as "complex roots only "