Question

For what values of x is the expression $15+4x-3{{x}^{2}}$ be negative?

Answer 1

Hint: First, before proceeding with this, we must know the method to solve the quadratic function so that we can get factors from it, and then we can compute the range for which the function gives the negative value. Then, for the above condition, we need the value of the function given which is $15+4x-3{{x}^{2}}$ must be less than zero. Then, solve for each of the factors separately so that we get a range for the function to get a negative value to the range of x for which the function is negative.

Complete step-by-step solution:
In this question, we are supposed to find the values of the x for the function $15+4x-3{{x}^{2}}$ must be negative.
So, before proceeding for this, we must know the method to solve the quadratic function so that we can get factors from it, and then we can compute the range for which the function gives the negative value.
As we want the final value of the given function gives a negative value after substitution of values of x in the equation given.
So, for the above condition, we need the value of the function given which is $15+4x-3{{x}^{2}}$ must be less than zero.
Now, equating the function less than zero we get:
$15+4x-3{{x}^{2}}<0$
Then, we bifurcate the term 4x as (9x-5x) so that we can take common factors from it as:
$15+9x-5x-3{{x}^{2}}<0$
Then, by taking common factors from the above equation as:
$\begin{align}
  & 3\left( 5+3x \right)-x\left( 5+3x \right)<0 \\
 & \Rightarrow \left( 3x+5 \right)\left( 3-x \right)<0 \\
\end{align}$
Now, we will use the wavy curve method in which the above factors are equated to zero to get the critical points where the function is always zero as:
$\left( 3x+5 \right)\left( 3-x \right)=0$
So, after solving, we get two critical points as:
$\begin{align}
  & \left( 3x+5 \right)=0 \\
 & \Rightarrow 3x=-5 \\
 & \Rightarrow x=\dfrac{-5}{3} \\
\end{align}$ and $\begin{align}
  & \left( 3-x \right)=0 \\
 & \Rightarrow x=3 \\
\end{align}$
So, we get the critical points and then we plot them on the number line and then check all the three ranges as:

So, we get three ranges as \[\left( -\infty ,\dfrac{-5}{3} \right)\], \[\left( \dfrac{-5}{3},3 \right)\]and \[\left( 3,\infty \right)\].
Now, we will consider the test point in each range to get the result of that range as starting with first range as \[\left( -\infty ,\dfrac{-5}{3} \right)\]and test point be -2, we get:
$\begin{align}
  & 15+4\left( -2 \right)-3{{\left( -2 \right)}^{2}} \\
 & \Rightarrow 15-8-12 \\
 & \Rightarrow -5 \\
\end{align}$
So, we get the result as negative which shows in the range \[\left( -\infty ,\dfrac{-5}{3} \right)\], equation $15+4x-3{{x}^{2}}$is negative.
Now, we will consider the test point in each range to get the result of that range as starting with first range as \[\left( \dfrac{-5}{3},3 \right)\]and test point be 0, we get:
$\begin{align}
  & 15+4\left( 0 \right)-3{{\left( 0 \right)}^{2}} \\
 & \Rightarrow 15+0-0 \\
 & \Rightarrow 15 \\
\end{align}$
So, we get the result as positive which shows in the range \[\left( \dfrac{-5}{3},3 \right)\], equation $15+4x-3{{x}^{2}}$is positive.
Now, we will consider the test point in each range to get the result of that range as starting with first range as \[\left( 3,\infty \right)\]and test point be 4, we get:
$\begin{align}
  & 15+4\left( 4 \right)-3{{\left( 4 \right)}^{2}} \\
 & \Rightarrow 15+16-48 \\
 & \Rightarrow -17 \\
\end{align}$
So, we get the result as positive which shows in the range \[\left( 3,\infty \right)\], equation $15+4x-3{{x}^{2}}$is positive.
So, by combining the above two range we get the final answer where $\cup $ is the symbol of union which represents the combination of two ranges as:
$\left( -\infty ,\dfrac{-5}{3} \right)\cup \left( 3,\infty \right)$
Hence, the value of x for which the function $15+4x-3{{x}^{2}}$ be negative is $\left( -\infty ,\dfrac{-5}{3} \right)\cup \left( 3,\infty \right)$.

Note: Now, to solve this type of question which has the quadratic equation of the form as $a{{x}^{2}}+bx+c=0$, we can also use the quadratic formula to calculate the roots required to get the range. So, the quadratic formula for the above equation is:
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
So, we get the same result as above.