<p>Hey everyone! Hope you are having a good day. I needed help with these two problems I got wrong on the 2012-2013 "Preparing for the ACT" math section. Thanks so much!</p>
<ol>
<li>In the equation </li>
</ol>
<p>x^2 + mx + n = 0, m and n are integers.
The only possible value for x is 3. What is the value
of m ?
A. 3
B. 3
C. 6
D. 6
E. 9
60. The solution set of which of the following equations is
the set of real numbers that are 5 units from −3 ?
F. abs (x + 3) = 5
G. abs(x − 3) = 5
H. abs(x + 5) = 3
J. abs(x − 5) = 3
K. abs(x + 5) = −3</p>
<p>x^2 + mx + n = 0, m and n are integers.
The only possible value for x is 3. What is the value
of m ?
A. 3
B. 3
C. 6
D. 6
E. 9</p>
<p>So, (-3)^2+m(-3)-n=0 (Substitute x for -3)
then 9-3m-0=0(0 is an integer so set n equal to 0)
9-3m=0
9=3m
m=3
So the answer is A</p>
<ol>
<li>The solution set of which of the following equations is
the set of real numbers that are 5 units from −3 ?
F. abs (x + 3) = 5
G. abs(x − 3) = 5
H. abs(x + 5) = 3
J. abs(x − 5) = 3
K. abs(x + 5) = −3</li>
</ol>
<p>I am guessing it’s F if I understand this question right. It’s basically asking which of the following solutions to abs value equations is + or - 5 from -3.</p>
<p>It’s F because |x+3|=5
so x+3=5 or x+3=-5
so x=5-3 or x= -3-5</p>
<p>In x=5-3 your adding 5 to -3 (5 units to the right from -3)
In x=-3-5 you subtracting 5 from -3 (5 units to the left from -3)</p>
<p>The equation you assumed the function to be is x^2+3x=0, which has 2 solutions for x, x=0 and x = -3.</p>
<p>My take on the question is this: The key is that they mentioned there is only one possible value for x, which means that the discriminant b^2-4ac = 0, or in this case, m^2-4n=0.</p>
<p>Then if you solve you have m^2=4n. This gets a tad bit tricky as both 6 and -6 can work for m, but because we are only solving for ONE solution, we have to check both solutions.</p>
<p>As such, we find that if m=-6, then x can only be 3, whereas if m=6 x can only be -3.</p>
<p>Therefore I believe the answer is C (6). Let me know if I’m wrong lol its been a while</p>
<p>Note that all quadratics have two solutions in the complex plane, and that any n-degree polynomial can be factored as a product of n linear polynomials.</p>
<p>The only solution is x = -3, so it turns out that the polynomial must be (x-(-3))(x-(-3)) = (x+3)(x+3) (with leading coefficient 1). Expanding, we get x^2 + 6x + 9, m = 6, C.</p>