Question

If the roots of the quadratic equation $k{{x}^{2}}+\left( a+b \right)x+ab$ are $-1,-b$, find the value of $k$?

Answer 1

Hint: We first take the general formula of roots for quadratic equation. We apply the roots $-1,-b$ for the equation $k{{x}^{2}}+\left( a+b \right)x+ab$. Then we use the relation of the ratio of roots being equal to find the value of $k$.

Complete step-by-step solution:
We know that for quadratic equation $a{{x}^{2}}+bx+c=0$, if the roots are $m,n$ then we can say that
$m+n=-\dfrac{b}{a}$ and $mn=\dfrac{c}{a}$.
Let's use the given roots $-1,-b$ for the equation $k{{x}^{2}}+\left( a+b \right)x+ab$.
Therefore, $-1-b=-\dfrac{a+b}{k}$ and $\dfrac{ab}{k}=b$.
Also, we know that the roots $-1,-b$ will satisfy the equation $k{{x}^{2}}+\left( a+b \right)x+ab$.
We can put the value in the equation to find the final value of the equation as 0.
We put $x=-1$ in $k{{x}^{2}}+\left( a+b \right)x+ab=0$.
So, $k{{\left( -1 \right)}^{2}}+\left( a+b \right)\left( -1 \right)+ab=0$. Simplifying we get
\[\begin{align}
  & k{{\left( -1 \right)}^{2}}+\left( a+b \right)\left( -1 \right)+ab=0 \\
 & \Rightarrow k-a-b+ab=0 \\
 & \Rightarrow k=a+b-ab \\
\end{align}\]
Now, we put $x=-b$ in $k{{x}^{2}}+\left( a+b \right)x+ab=0$.
So, $k{{\left( -b \right)}^{2}}+\left( a+b \right)\left( -b \right)+ab=0$. Simplifying we get
\[\begin{align}
  & k{{\left( -b \right)}^{2}}+\left( a+b \right)\left( -b \right)+ab=0 \\
 & \Rightarrow k{{b}^{2}}-ab-{{b}^{2}}+ab=0 \\
 & \Rightarrow k{{b}^{2}}={{b}^{2}} \\
 & \Rightarrow k=\dfrac{{{b}^{2}}}{{{b}^{2}}}=1 \\
\end{align}\]
The value of $k$ is 1.

Note: We need to remember that we can also solve the equations $-1-b=-\dfrac{a+b}{k}$ and $\dfrac{ab}{k}=b$ to find the value of $a$ and $b$ to solve the problem. The equation becomes ${{x}^{2}}+\left( a+b \right)x+ab$ which also gives the value of $a$ and $b$.