<p>Can someone help me solve the following problem?</p>
<p>In the equation x^2 + mx + n = 0, m and n are integers. The only possible value for x is -3. What is the value of m?</p>
<p>There is not enough information to solve this problem…</p>
<p>There isn’t enough info. This is the farthest that you can get with this equation:</p>
<p>x^2 + mx + n = 0
9+(-3)m+n=0
(-3m)+n=-9
m+n=3</p>
<p>Then, if m and n are integers, m could equal 0, 1, 2, or 3.</p>
<p>@Cant
You kinda divided by -3 in your problem but didn’t divide the n by -3 with the rest of the equation.</p>
<p>@All
Integers can be negative too.</p>
<p>@xplosneer you dont need to divide the n by -3, you’re only dividing the -3 from the m, and since you did it to the left side, you must do it to the right side and divide -9 by -3.</p>
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<p>x1 = (- m ± root of (m^2 - 4<em>1</em>n)) / 2</p>
<p>-3 = ( - m ± root of(m^2 - 4n))/ 2</p>
<p>-m ± root of (m^2 - 4n) = - 6</p>
<p>± root of(m^2 - 4n) = -6 + m</p>
<p>BY inspection as m,n are integers:</p>
<p>m = 12, n = 27</p>
<p>i.e. root of(144 - 108) = -6 + 12</p>
<p>6 = 6</p>
<p>Checking:</p>
<p>-3^2 - 3*12 + 27 = 0 Yes</p>
<p>So m = 12</p>
<p>@hero</p>
<p>Did I miss something? What you are proposing is true only for adding/subtracting to each side.</p>
<p>You can’t only divide single terms in an equation unless you take the term out of each part of the equation.
That’s like saying 3x+y=13. If x=3 and y=4, then 9+4=13 but also 3+4=(13/3).</p>
<p>Haha woops. I guess this is what happens when I’ve been out of school for 2 months…</p>
<p>lol god I’m so stupid this is so easy. Since there is only one root.</p>
<p>If x can only be -3, then:</p>
<p>(x+a)^2 = 0 so (-3 + a)^2 = 0, a = 3
(x-a)^2 = 0 so (-3 - a)^2 = 0, a = -3</p>
<p>So x^2 - 2xa + a^2 = 0, or x^2 + 2xa + a^2 = 0</p>
<p>So x^2 + 6x + 9 = 0 (Using both equations with a = 3, a = -3)</p>
<p>-3^2 + 6(-3) + 9 = 9 -18 + 9 = 0</p>
<p>So m = -2a or m = 2a and n = a^2</p>
<p>m = -2(-3) = 6 and m = 2a = 2(3) = 6, n = -3^2 = 3^2 = 9</p>
<p>So m = 6 and n = 9</p>
<p>since we are on the topic of solving math problems, can someone pleaseeee help me with this one?!?</p>
<ol>
<li>Paul randomly selects 3 apples from a basket containing 4 green apples and 6 red apples. What is the probability that he selects at least 1 green apple?</li>
</ol>
<p>(A) 1/6
(B) 3/10
(C) 5/12
(D) 5/6
(E) 29/33</p>
<p>I just don’t get probabilities</p>
<p>You can select 1 green apple from three choices (3,1) 3 different ways.</p>
<p>For zero green apples: p(x) = (3/5)^3 = 27/125
for one: p(x) = 3<em>(2/5)</em>(3/5)^2 = 54/125
for two: p(x) = 3<em>(2/5)^2</em>(3/5) = 36/125
for three: p(x) = (2/5)^3 = 8/125</p>
<p>1 + 2 + 3 green apples = 98/125</p>
<p>Basically if 3 is the ONLY solution then (x-3)(x-3)=0,
FOIL: x^2+6x+9=0
Therefore m=6</p>
<p>@jamiegirl best is to find the probability of finding P(zero green apples) and subtract from 1, since that is the complementary event. P(zero green apples) = (6C3)/(10C3) = 20/120 = 1/6 (since we want to select three red apples - @jsanche32 's solution assumes replacement, which I highly doubt is correct), so the probability that Paul selects at least one green apple is 1 - (1/6) = (D) 5/6.</p>
<p>(x+3)(x+3)
x^2+6x+9
6</p>