Question

If the system of equations \[kx-5y=2,6x+2y=7\] has no solution, find the quadratic equation whose roots are k – 1 and k + 1.

Answer 1

Hint: Solve the two given equations by elimination variable y. Find the value of x in terms of k. Substitute the denominator of the expression of x obtained equal to 0 and find the value of k. Assume the roots of the required equation as ‘a’ and ‘b’. Find the values of ‘a’ and ‘b’ by substituting the obtained value of k in (k -1) and (k + 1) respectively. Finally, write the quadratic equation as: - \[{{x}^{2}}-\left( a+b \right)x+ab\].

Complete step-by-step solution
Here, we have been provided with two equations: - \[kx-5y=2\] and \[6x+2y=7\] which has no solution. We have to find a quadratic equation whose roots are k – 1 and k + 1. First, let us find the value of k.
\[\Rightarrow kx-5y=2\] - (1)
\[\Rightarrow 6x+2y=7\] - (2)
Multiplying equation (1) with 2 and equation (2) with 5, we get,
\[\Rightarrow 2kx-10y=4\] - (3)
\[\Rightarrow 30x+10y=35\] - (4)
Adding equations (3) and (4), we get,
\[\begin{align}
  & \Rightarrow \left( 2k+30 \right)x=39 \\
 & \Rightarrow x=\dfrac{39}{2k+30} \\
\end{align}\]
Now, since the two equations have no solution therefore the value of x must not be real. This will happen when the denominator of the above expression becomes zero. So, we have,
\[\begin{align}
  & \Rightarrow 2k+30=0 \\
 & \Rightarrow 2k=-30 \\
 & \Rightarrow k=-15 \\
\end{align}\]
Now, it is given that the roots of the quadratic equation are: - (k - 1) and (k + 1). Let us assume the two roots as ‘a’ and ‘b’.
\[\begin{align}
  & \Rightarrow a=k-1=-15-1=-16 \\
 & \Rightarrow b=k+1=-15+1=-14 \\
\end{align}\]
Now, we know that a quadratic equation having two roots as ‘a’ and ‘b’ is given as: - \[{{x}^{2}}-\left( a+b \right)x+ab=0\]. Therefore, substituting the values of ‘a’ and ‘b’, we get,
\[\Rightarrow \] Required quadratic equation = \[{{x}^{2}}-\left( -16+\left( -14 \right) \right)x+\left( -16 \right)\times \left( -14 \right)=0\]
 \[\Rightarrow \] Required quadratic equation = \[{{x}^{2}}+30x+224=0\]
Hence, the required quadratic equation is given as: - \[{{x}^{2}}+30x+224=0\]

Note: One may note that we have used the elimination method to solve the two given equations and then find the value of k. One can also use different methods to find the value of k. As the given equations contains two variables, so write the two equations in the form: - \[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0\] and \[{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0\]. Now, apply the condition of system of equations having no solution given as: - \[\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}\ne \dfrac{{{c}_{1}}}{{{c}_{2}}}\] and find the value of k by considering \[\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}\].