Question

What is the product of 2 solutions of the equation ${{x}^{2}}+3x-21=0$?

Answer 1

Hint: We solve this problem by finding the solutions of the given equation and then multiply them to get the required product. We use the formula to find the solutions of quadratic equation in the form of $a{{x}^{2}}+bx+c=0$ are given as,
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
By using this formula we find the roots of a given equation and then find the product.

Complete step-by-step answer:
We are given with the equation as,
${{x}^{2}}+3x-21=0$
Here, we can see that the above equation is quadratic equation.
Now, let us use the formula to find the solutions of quadratic equations.
We know that the solution of quadratic equation of the form $a{{x}^{2}}+bx+c=0$ are given as,
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Now, by comparing the both general form of quadratic equation and given equation then we get,
$\begin{align}
  & \Rightarrow a=1 \\
 & \Rightarrow b=3 \\
 & \Rightarrow c=-21 \\
\end{align}$
By using these values in the above roots formula then we get,
$\begin{align}
  & \Rightarrow x=\dfrac{-3\pm \sqrt{{{3}^{2}}-4\left( 1 \right)\left( -21 \right)}}{2\left( 1 \right)} \\
 & \Rightarrow x=\dfrac{-3\pm \sqrt{9+84}}{2} \\
 & \Rightarrow x=\dfrac{-3\pm \sqrt{93}}{2} \\
\end{align}$
Now, let us take the two solutions from the above equation as,
$\begin{align}
  & \Rightarrow \alpha =\dfrac{-3+\sqrt{93}}{2} \\
 & \Rightarrow \beta =\dfrac{-3-\sqrt{93}}{2} \\
\end{align}$
Now, let us find the product of these two solutions then we get,
$\begin{align}
  & \Rightarrow \alpha \times \beta =\left( \dfrac{-3+\sqrt{93}}{2} \right)\times \left( \dfrac{-3-\sqrt{93}}{2} \right) \\
 & \Rightarrow \alpha \times \beta =\dfrac{\left( -3+\sqrt{93} \right)\times \left( -3-\sqrt{93} \right)}{4} \\
\end{align}$
We know that the standard formula of algebra that is,
$\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$
By using this formula in above equation then we get,
$\begin{align}
  & \Rightarrow \alpha \times \beta =\dfrac{{{\left( -3 \right)}^{2}}-{{\left( \sqrt{93} \right)}^{2}}}{4} \\
 & \Rightarrow \alpha \times \beta =\dfrac{9-93}{4} \\
 & \Rightarrow \alpha \times \beta =\dfrac{-84}{4} \\
\end{align}$
Here, we know that the quotient when 84 is divided by 4 is 21. By using this result in above equation we get,
$\Rightarrow \alpha \times \beta =-21$
Therefore, we can conclude that the product of two solutions of ${{x}^{2}}+3x-21=0$ is $'-21'$

Note: We have a shortcut for finding the required answer without finding the roots.
We have the standard result that is $\alpha ,\beta $ are the roots of standard quadratic equation $a{{x}^{2}}+bx+c=0$ then,
(1) Sum of roots $\alpha +\beta =\dfrac{-b}{a}$
(2) Product of roots $\alpha \times \beta =\dfrac{c}{a}$
By using the above result to the given equation ${{x}^{2}}+3x-21=0$ we get the required product of two roots as,
$\begin{align}
  & \Rightarrow \alpha \times \beta =\dfrac{-21}{1} \\
 & \Rightarrow \alpha \times \beta =-21 \\
\end{align}$
Therefore, we can conclude that the product of two solutions of ${{x}^{2}}+3x-21=0$ is $'-21'$