<p>If x and y are real numbers such that x > 1 and y < -1, then which of the following inequalities must be true?</p>
<p>Why is it
C) x/3 - 5 > y/3 - 5</p>
<p>and not..</p>
<p>E) x^-2 > y^-2</p>
<p>I had C originally, but as i went through, E seemed always true.</p>
<p>Also, how does one solve this question?</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>A. 3
B. -3
C. 6
D. -6
E. 9</p>
<p>Thanks.</p>
<p>
</p>
<p>so e)</p>
<p>1/x^2 > 1/y^2 … Well, anything squared is always positive so, if |x|=|y|, then 1/x^2 = 1/y^2 (e.g, x=2, y = -2) , which disproves the inequality. if |x| > |y| then the inequality is once again disproved (e.g x = 10, y = -4).</p>
<p>c) is always true.</p>
<p>
</p>
<p>Okay, so only when x is equal to -3 is the equation satisfied so that means:</p>
<p>(x+3)(x+3) = x^2 + mx + n </p>
<p>mutiply it out:</p>
<p>x^2 + 6x + 9 = x^2 + mx + n</p>
<p>mx matches with the 6x term, so m = 6. (c)</p>
<p>Thanks a lot. I really do not know how I had trouble with the first question… that should have been basic math for me. lol. oh well. Thanks again.</p>
<p>Hi,
Please help to explain why (x+3)(x+3) = x^2 + mx + n is satisfied</p>
<p>when x is equal to -3?</p>
<p>how do you derive x+3) (x+3)? thank you!</p>