Question

How do you solve \[14-5x-{{x}^{2}}<0\]?

Answer 1

Hint: The given problem is an inequality having a quadratic equation on its left side. To solve these types of questions, we need to follow some steps. Let \[q(x)\] is a quadratic equation, and we need to solve \[q(x)<0\]. As this is an inequality it will give a range of values as its solution. To find the range, we need to find the roots of the quadratic equation \[q(x)\]. Let \[\alpha \And \beta \] be the roots of the equation, and \[\alpha \] is the smaller root. Then the solution range of the equation is \[\left( \alpha ,\beta \right)\].But, if the coefficient of the square term is negative then range is \[\left( -\infty ,\alpha \right)\bigcup \left( \beta ,\infty \right)\]. To find the roots, we can use the formula method.

Complete step by step solution:
We are asked to solve the inequality \[14-5x-{{x}^{2}}<0\]. Rearranging the terms in the left side, it can be written as
\[\Rightarrow -{{x}^{2}}-5x+14<0\]
To find the roots, we will use the formula method. Thus, the roots of the equation are
\[\begin{align}
  & \Rightarrow x=\dfrac{-\left( -5 \right)\pm \sqrt{{{\left( -5 \right)}^{2}}-4\left( -1 \right)(14)}}{2\left( -1 \right)} \\
 & \Rightarrow x=\dfrac{5\pm \sqrt{25+56}}{-2} \\
 & \Rightarrow x=\dfrac{5\pm 9}{-2} \\
\end{align}\]
\[\Rightarrow x=\alpha =\dfrac{5+9}{-2}\] or \[x=\beta =\dfrac{5-9}{-2}\]
\[\therefore \alpha =-7\] or \[\beta =2\]
Here, the coefficient of a square term is negative, hence the solution range will be \[\left( -\infty ,-7 \right)\bigcup \left( 2,\infty \right)\].
Hence, the given inequality will hold for the range \[\left( -\infty ,-7 \right)\bigcup \left( 2,\infty \right)\].

Note:
Here, the inequality sign was less than \['<'\] so the solution range was of the form \[\left( \alpha ,\beta \right)\]. For an inequality of the form \[q(x)>0\], here \[q(x)\] is a quadratic equation. The solution range is of the form \[\left( -\infty ,\alpha \right)\bigcup \left( \beta ,\infty \right)\], \[\alpha \And \beta \]are the roots of the equation, and \[\alpha \] is the smaller root. If the inequality signs are \[\le /\ge \], then we have to take the closed intervals for the roots.