Question

The roots $\alpha $ and $\beta $ of the quadratic equation ${{x}^{2}}-5x+3\left( k-1 \right)=0$ are such that $\alpha -\beta =11$ . Find the value of k

Answer 1

Hint: To find the value of k, we have to find $\alpha $ and $\beta $ . We know that for a quadratic polynomial $a{{x}^{2}}+bx+c$ , the sum of the roots, $\alpha +\beta =\dfrac{-b}{a}$ . We have to compare the standard form to the given polynomial and substitute the values in the equation. We are also given that $\alpha -\beta =11$ . We have to solve these two equations and find the value of $\alpha $ and $\beta $ . For a quadratic polynomial, the product of the roots is equal to $\dfrac{c}{a}$ . We have to compare the standard form to the given polynomial, substitute the values in the equation and solve for k.

Complete step by step solution:
We are given that ${{x}^{2}}-5x+3\left( k-1 \right)=0$ . We know that for a quadratic polynomial $a{{x}^{2}}+bx+c$ , the sum of the roots of this polynomial is $\dfrac{-b}{a}$ . Let $\alpha $ and $\beta $ be the roots.
$\Rightarrow \alpha +\beta =\dfrac{-b}{a}$
From the given equation, we can see that $a=1,b=-5$ and $c=3\left( k-1 \right)$ . Therefore, we can write the sum of the roots as
$\begin{align}
  & \Rightarrow \alpha +\beta =\dfrac{-\left( 5 \right)}{1} \\
 & \Rightarrow \alpha +\beta =5...\left( i \right) \\
\end{align}$
We know that the product of the roots is equal to $\dfrac{c}{a}$ .
$\Rightarrow \alpha \beta =\dfrac{c}{a}$
From the given equation, we know that $a=1\text{ and }c=3\left( k-1 \right)$ . Therefore, we can write the product of the roots as
$\begin{align}
  & \Rightarrow \alpha \beta =\dfrac{3\left( k-1 \right)}{1} \\
 & \Rightarrow \alpha \beta =3\left( k-1 \right)...\left( ii \right) \\
\end{align}$
We are given that $\alpha -\beta =11...\left( iii \right)$
Let us add equations (i) and (iii).
$\Rightarrow \alpha +\beta +\alpha -\beta =5+11$
We can see that $\beta $ cancels out.
\[\begin{align}
  & \Rightarrow \alpha +\require{cancel}\cancel{\beta }+\alpha -\require{cancel}\cancel{\beta }=5+11 \\
 & \Rightarrow 2\alpha =16 \\
\end{align}\]
We have to take 2 to the RHS.
$\begin{align}
  & \Rightarrow \alpha =\dfrac{16}{2} \\
 & \Rightarrow \alpha =8 \\
\end{align}$
Let us substitute the above value in equation (i).
$\begin{align}
  & \Rightarrow 8+\beta =5 \\
 & \Rightarrow \beta =5-8 \\
 & \Rightarrow \beta =-3 \\
\end{align}$
Now, we have substitute the values of $\alpha $ and $\beta $ in equation (ii).
$\begin{align}
  & \Rightarrow 8\times -3=3\left( k-1 \right) \\
 & \Rightarrow -24=3\left( k-1 \right) \\
\end{align}$
We have to apply distributive property on the RHS.
$\Rightarrow -24=3k-3$
Let us take -3 to the LHS.
$\begin{align}
  & \Rightarrow -24+3=3k \\
 & \Rightarrow 3k=-24+3 \\
 & \Rightarrow 3k=-21 \\
\end{align}$
We have to take 3 to the RHS.
$\begin{align}
  & \Rightarrow k=\dfrac{-21}{3} \\
 & \Rightarrow k=-7 \\
\end{align}$
Hence, the value of k is -7.

Note: Students must know properties of the roots of quadratic polynomials. They must know to solve two equations. We can also find the value of k in an alternate method.
We have obtained three equations from the above solution.
$\begin{align}
  & \Rightarrow \alpha +\beta =5...\left( i \right) \\
 & \alpha \beta =3\left( k-1 \right)...\left( ii \right) \\
 & \alpha -\beta =11...\left( iii \right) \\
\end{align}$
We know that ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ . We can write $\left( a-b \right)$ from this formula as
$\left( a-b \right)=\sqrt{{{a}^{2}}-2ab+{{b}^{2}}}$
We have to substitute $a=\alpha $ and $b=\beta $ in the above formula.
$\Rightarrow \left( \alpha -\beta \right)=\sqrt{{{\alpha }^{2}}-2\alpha \beta +{{\beta }^{2}}}$
Let us add and subtract $2\alpha \beta $ to the RHS.
$\begin{align}
  & \Rightarrow \left( \alpha -\beta \right)=\sqrt{{{\alpha }^{2}}+{{\beta }^{2}}+2\alpha \beta -2\alpha \beta -2\alpha \beta } \\
 & \Rightarrow \left( \alpha -\beta \right)=\sqrt{\left( {{\alpha }^{2}}+{{\beta }^{2}}+2\alpha \beta \right)-4\alpha \beta } \\
\end{align}$
We know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ . Therefore, the above equation becomes
$\Rightarrow \left( \alpha -\beta \right)=\sqrt{{{\left( \alpha +\beta \right)}^{2}}-4\alpha \beta }$
Let us substitute equations (i), (ii) and (iii) in the above equation.
$\begin{align}
  & \Rightarrow 11=\sqrt{{{5}^{2}}-4\times 3\left( k-1 \right)} \\
 & \Rightarrow 11=\sqrt{25-12k+12} \\
 & \Rightarrow 11=\sqrt{37-12k} \\
\end{align}$
Let us square both the sides.
$\begin{align}
  & \Rightarrow {{11}^{2}}=37-12k \\
 & \Rightarrow 121=37-12k \\
 & \Rightarrow 121-37=-12k \\
 & \Rightarrow 84=-12k \\
 & \Rightarrow -12k=84 \\
\end{align}$
We have to take -12 to the RHS.
$\begin{align}
  & \Rightarrow k=\dfrac{84}{-12} \\
 & \Rightarrow k=-7 \\
\end{align}$