<p>Someone please help me answer this SAT math question!?
If (c+1)^2 = -b, which of the following statements could be true?</p>
<ol>
<li>c>0</li>
<li>c=0</li>
<li>c<0</li>
</ol>
<p>A) none
B) 1 only
C) 3 only
D) 1 and 2 only
E) 1, 2, and 3</p>
<p>Please explain how you get ur answer!</p>
<p>I got A) none. is this right?</p>
<p>@diddly123 that’s what I also thought. (c+1)^2 simplifies to c^2 + 1 = -b. You can’t square a number and get a negative product.</p>
<p>same thinking!</p>
<p>(c+1)^2 is not equal to c^2 + 1! </p>
<p>(c+1)^2=(c+1)(c+1)=c^2+2c+1.</p>
<p>You do not want to multiply this out for this problem however.</p>
<p>Here is a solution:</p>
<p>c=-1, b=0 works.</p>
<p>c=0, b=-1 works</p>
<p>c=1, b=-4 works.</p>
<p>So the answer is choice (E).</p>
<p>@DrSteve, oops, dumb moment. Haha, sorry about that.</p>
<p>@Yakisoba. </p>
<p>It’s actually nice that you wrote that down. It’s a very common mistake so it’s great to bring that standard trap to students’ attention.</p>
<p>This problem relies on a common misconception that -b must be a negative number. What if b itself is negative, in that case -b is positive. </p>
<p>When I ask students about solution to the following equation:
x^2 = -x
a lot of them say x^2 is always positive, so how can it be equal to a negative number? </p>
<p>However, x=-1 and x=0 are the two possible solutions. </p>
<p>A similar trap is presented in the above problem.</p>
<p>So it would only be A if the question read “(c+1)^2=b, where b is a negative number”?</p>
<p>Yes, ybrown: (c+1)^2 can never be negative, no matter the value of c.</p>
<p>Classic SAT problem. So many kids will be in a hurry and will see the square power and the -b and pick A because they know anything squared can’t be negative. Without any restrictions on what ‘b’ can be, there will be an infinite set of solutions.</p>
<p>Agree with everyone, answer: (E)</p>
<p>oops! let’s just forget i missed that lol. yeah i was confused though cause it didn’t say anything about b so i just assumed it meant negative number. gotta watch out for that!</p>
<p>(c+1)^2=-b
=> c^2+2c+1=-b
=> c^2+2c+(1+b)=0
Since b and c is real number (assume it is since the question does not say is not), its discriminant must be greater or equal to 0
4c^2-4c^2(1+b)>or=0
=> 4c^2[1-(1+b)]>or=0
=> -4bc^2>or=0
Now, we have two things to discuss(I think is pretty clear that b may be 0 if c=-1)
1)b>0
2)b<0</p>
<p>1)If b>0, then c=0
2)If b<0, then c is any real number</p>
<p>Therefore, the answer is (E)</p>
<p>Since we already know the c is any real number, we do not have to consider c as a imaginary number.</p>