Question

Find the values of \[m\] for which \[\left( {m - 2} \right){x^2} + 8x + m + 4 > 0\] for all real \[x\].

Answer 1

Hint: Here, in the given question, we are asked to find the values of \[m\] for which \[\left( {m - 2} \right){x^2} + 8x + m + 4 > 0\] for all real \[x\]. The demand of the question is that the given quadratic polynomial should be greater than zero or we can say it should be positive. To solve the question, we must know about the conditions which are required to satisfy the given problem. Any quadratic polynomial is positive when its leading coefficient is positive and its discriminant is negative.

Complete step-by-step solution:
Given quadratic polynomial,
\[\left( {m - 2} \right){x^2} + 8x + m + 4\]
Now, we know that, any quadratic function \[a{x^2} + bx + c\] is positive only and only when it satisfies the two conditions which are given as:
\[\left( i \right)a > 0 \\
  \left( {ii} \right){b^2} - 4ac < 0 \]
If we compare the given quadratic polynomial with the standard form of quadratic polynomial, we get,
\[ a = \left( {m - 2} \right) \\
  b = 8 \\
  c = m + 4 \]
It means if \[\left( {m - 2} \right){x^2} + 8x + m + 4\] should be greater than zero, then, two conditions must be satisfied:
\[\left( {m - 2} \right) > 0\]and\[{b^2} - 4ac < 0\]
If \[\left( {m - 2} \right) > 0\]
\[ \Rightarrow m > 2\]
For \[{b^2} - 4ac < 0\],
\[{\left( 8 \right)^2} - 4\left( {m - 2} \right)\left( {m + 4} \right) < 0\]
Simplifying it,
\[ 64 - 4\left( {{m^2} + 4m - 2m - 8} \right) < 0 \\
   \Rightarrow 64 - 4{m^2} - 8m + 32 < 0 \\
   \Rightarrow 96 - 4{m^2} - 8m < 0 \]
Multiplying by \[\left( { - 1} \right)\] both sides and rearranging it, we get,
\[4{m^2} + 8m - 96 > 0\]
Dividing by\[4\]both sides, we get,
\[{m^2} + 2m - 24 > 0\]
Now again this can be written as,
\[{m^2} + 6m - 4m - 24 > 0\]
Taking \[m\] common from the first two terms and \[ - 4\] common from last two terms, we get,
\[m\left( {m + 6} \right) - 4\left( {m + 6} \right) > 0\]
Taking \[\left( {m + 6} \right)\] common,
\[\left( {m + 6} \right)\left( {m - 4} \right) > 0\]
Product of two numbers can be positive in two cases, either both the numbers should be positive or both the numbers should be negative.
If \[\left( {m + 6} \right),\left( {m - 4} \right)\] are positive,
Then, \[
  m + 6 > 0 \\
   \Rightarrow m > - 6 \]
And, \[
  m - 4 > 0 \\
   \Rightarrow m > 4 \]
If both the numbers are positive, the value of \[m\] for which \[\left( {m + 6} \right)\left( {m - 4} \right) > 0\] holds true will be \[\left( {m > - 6} \right) \cap \left( {m > 4} \right) = \left( {m > 4} \right)\]
If \[\left( {m + 6} \right),\left( {m - 4} \right)\] are negative,
Then,
 \[m + 6 < 0 \\
 \Rightarrow m < - 6 \]
And,
\[m - 4 < 0 \\
   \Rightarrow m < 4 \]
If both the numbers are negative, the value of \[m\] for which \[\left( {m + 6} \right)\left( {m - 4} \right) > 0\] holds true will be
\[\left( {m < - 6} \right) \cap \left( {m < 4} \right) = \left( {m < - 6} \right)\]
But from the first condition, we know that \[m > 2\]
Hence, the value of \[m\] for which \[\left( {m - 2} \right){x^2} + 8x + m + 4 > 0\] will be
\[\left( {m > 2} \right) \cap \left[ {\left( {m > 4} \right) \cup \left( {m < - 6} \right)} \right]
   = m > 4 \]

Note: This question needs to be solved very carefully. It is very important to understand the concepts of union and intersection, otherwise we can make a mistake in selecting a region for \[m\]. Any minor mistake can lead to a wrong answer at the end. Or if anyone is still confused, then he/she must check the solution by putting back into the given function.