<p>If X^2 + 8x ≤ -15, which of the following is a possible value of x? </p>
<p>A) -6
B) -4 </p>
<p>I put A but it's B
My working </p>
<p>X^2 + 8x + 15 ≤ 0</p>
<p>(x+5)(x+3) ≤ 0</p>
<p>x+5 ≤ 0 therefore x ≤ -5 or x+3 ≤ 0 therefore x ≤ -3 </p>
<p>I find this question strange because the answers could be A or B </p>
<p>Help much appreciated
cheers</p>
<p>Put -6 into your equation. You may be surprised :)</p>
<p>wait how is that possible!!! when the math says that it can either be -6 or -4. What on earth is going on?!</p>
<p>You have a conceptual misunderstanding. Here’s what the issue is:</p>
<p>If AB=0 then A =0 or B =0 or both.</p>
<p>But if AB< 0, then you can not say that either A<0 or B<0 or BOTH!
You need one to be positive and the other negative.</p>
<p>So if (x+3)(x+5)<0, x has to be BETWEEN -3 and -5…</p>
<p>Here is another variation on the theme:</p>
<p>Let w=(2-x)(5-x)(8-x)(12-x)</p>
<p>For how many integer values of x is w less than zero?</p>
<p>ahh right okay thanks for that then.</p>