Am I doing Vieta's Formulas correctly?I've been given this exercise: x2−(m+3)x+m+2, I'm supposed to check for which real values of the parameter m can be used for this to work:1x1+1x2&gt;12 (I multiply both sides by 2x1x2) and get the following:2(x1+x2)−x1x2&gt;0which results to m≻4 using Vieta's formulasAfterwards I have another argument, x12+x22&lt;5After solving I get that m is in the interval (−4,0)My book tells me the final result for possible M solutions is in the interval (−2,0).

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Am I doing Vieta&#039;s Formulas correctly?I&#039;ve been given this exercise: x2−(m+3)x+m+2,  I&#039;m supposed to check for which real values of the parameter m can be used for this to work:1x1+1x2&amp;gt;12 (I multiply both sides by 2x1x2) and get the following:2(x1+x2)−x1x2&amp;gt;0which results to m≻4 using Vieta&#039;s formulasAfterwards I have another argument, x12+x22&amp;lt;5After solving I get that m is in the interval (−4,0)My book tells me the final result for possible M solutions is in the interval (−2,0).

massifyc9 · Accepted Answer

You can&#039;t multiply by&amp;nbsp;x1x2&amp;nbsp;since you don&#039;t know if it&#039;s a positive or negative quantity (remember the sign of the inequality would have to swap if it were negative, and stay the same otherwise).Remember what Viete&#039;s formulae tell you, that&amp;nbsp;x1+x2=m+3&amp;nbsp;and that&amp;nbsp;x1x2=m+2. You can use these if you simplify the left-hand side:1x1+1x2=x1+x2x1x2=m+3m+2,so you want to ensure that m is such thatm+3m+2&amp;gt;12.We can&#039;t multiply throughout by&amp;nbsp;m+2&amp;nbsp;since we don&#039;t know its sign. We can multiply by&amp;nbsp;(m+2)2, this is surely non-negative. This gives us(m+3)(m+2)&amp;gt;12m+22which simplifies to(m+2)(m+4)&amp;gt;0.A product of two numbers is&amp;nbsp;&amp;gt;0&amp;nbsp;either if they are both&amp;nbsp;&amp;gt;0, or if they are both&amp;nbsp;&amp;lt;0.In the first case (when&amp;nbsp;m+2&amp;nbsp;and&amp;nbsp;m+4&amp;nbsp;are both positive), we have&amp;nbsp;m≻2&amp;nbsp;and&amp;nbsp;m≻4, which is simply equivalent to saying&amp;nbsp;m≻2.In the second case (when they are both negative), we have&amp;nbsp;m&amp;lt;−2&amp;nbsp;and&amp;nbsp;m&amp;lt;−4, which is the same as saying that&amp;nbsp;m&amp;lt;−4.So in summary, your condition is equivalent to saying that&amp;nbsp;    ◻                            m          &amp;lt;          −          4                &amp;nbsp;or&amp;nbsp;                  m          &amp;gt;          −          2                      .

Avery Campbell · Accepted Answer

Step 11x1+1x2&amp;gt;12⇔2x2+2x1-x1x22x1x2&amp;gt;0⇔2(m+3)-m-22(m+2)=m+42(m+2)&amp;gt;0⇔(m+2)(m+4)&amp;gt;0⇔m∈(-∞,-4)∪(-2,∞)&amp;nbsp;Step 2x12+x22&amp;lt;5⇔(x1+x2)2-2x1x2&amp;lt;5⇔(m+3)2-2(m+2)-5=m(m+4)&amp;lt;0⇔-4&amp;lt;m&amp;lt;0Therefore&amp;nbsp;−2&amp;lt;m&amp;lt;0.

Am I doing Vieta's Formulas correctly? I've been given

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