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If \(m\) is a real number and \(x^2+mx+4\) has two distinct real roots, then what are the possible values of \(m\)? Express your answer in interval notation.

 

I've done a lot of quadratic arithmetic before, but I am stuck on this one. EP, can you take a shot at this one? I know you're good at tackling things like this.

 Mar 22, 2020
 #1
avatar+23710 
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For this to be true   the discriminat   b^2 - 4ac  > 0

 

m^2 -4(1)(4) > 0

m^2 > 16

 

m <-4       m >4

 

(-inf,-4) U (4, + inf)

 Mar 22, 2020
 #2
avatar+485 
+2

Hi TGO, I'm not EP, but I'll give it my best shot.

 

Often times with problems like these, you start by looking at the discriminant of given quadratic. The discriminant is defined in an equation \(ax^2+bx+c = 0\) as \(\sqrt{b^2-4ac}\). If the discriminant is negative, then the roots are imaginary, and if it is non negative, then the roots are real. Let's look at the discriminant of this equation:

Given the quadratic

\(x^2+mx+4\)

The discriminant would be:

\(\sqrt{m^2-16}\)

We can write the inequality:

\(\sqrt{m^2-16} \geq 0\)

Squaring both sides, we get:

\(m^2-16 \geq 0\)

This factors out to:

\((m+4) * (m-4) \geq 0\)

With these types of problems, it's best to create a sign chart!

Although I don't have the means to create one as of now, I will explain all the cases in this "chart". 

 

The first case is where: 

\(m < -4\)

If this is true, then the first term(m+4) would be negative, and so would the second term, meaning that a negative * negative = positive, meaning that when m < -4, the inequality holds true.

 

The second case is where:

\(-4\leq m \leq 4\).

Testing numbers in this range, it's clear that none of them work. If m is negative but still greater than -4, the first term is positive, but the second negative, making the entire expression negative. If m is positive but less than 4, the second term is negative, and the first term is positive, still leaving the expression unsatisfied. It's clear that this interval doesn't work. Also realize that m can't be -4 or 4, because that would make the discriminant 0, meaning there's a double root(two of the same root). However, the problem says that the roots must be distinct

 

The third case is where:

\(4 < m\)

In this case, both terms are left positive, satisfying the inequality. 

 

This means that there are two intervals which we have:

\((-\infty,-4) \cup (4,\infty)\)

.
 Mar 22, 2020
edited by jfan17  Mar 22, 2020
edited by jfan17  Mar 22, 2020
 #3
avatar+56 
+2

Thank you guys go much, EP and jfan17, especially you, jfan17 for having such an elaborate explanation.

laugh

 Mar 22, 2020
 #4
avatar+485 
+2

Any time!

jfan17  Mar 22, 2020
 #5
avatar+1970 
+2

Wow! That was one complicated problem!!! NICE!

CalTheGreat  Mar 22, 2020

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