Question

For what value of k, the equation \[2{{x}^{2}}-\left( k-1 \right)x+8=0\] has real and equal roots.

Answer 1

Hint: We know the standard form of the quadratic equation, \[a{{x}^{2}}+bx+c\] . Now, compare the standard form of the quadratic equation with the equation \[2{{x}^{2}}-\left( k-1 \right)x+8=0\] and get the values of a, b, and c. We know the formula for the discriminant of the quadratic equation, Discriminant = \[{{b}^{2}}-4ac\] . Use this formula and get the value of the discriminant. Now, for equal and real roots the discriminant of the quadratic equation must be equal to zero. Solve it further and get the value of k.

Complete step-by-step answer:
According to the question, we have the quadratic equation \[2{{x}^{2}}-\left( k-1 \right)x+8=0\] .
\[2{{x}^{2}}-\left( k-1 \right)x+8=0\] …………………………..(1)
We know the standard form of the quadratic equation, \[a{{x}^{2}}+bx+c\] …………………………(2)
Now, comparing equation (1) and equation (2), we get
\[a=2\] …………………….(3)
\[b=-\left( k-1 \right)\] …………………………..(4)
\[c=8\] …………………………..(5)
Here, we need to get the discriminant. The value of the discriminant for the quadratic equation \[a{{x}^{2}}+bx+c\] is given by the formula,
Discriminant = \[{{b}^{2}}-4ac\] …………………………..(6)
Now, putting the values of a from equation (3), b from equation (4), and c from equation (5) in the formula shown in equation (6), we get
Discriminant = \[{{\left\{ -\left( k-1 \right) \right\}}^{2}}-4\left( 2 \right)\left( 8 \right)={{k}^{2}}+1-2k-64={{k}^{2}}-2k-63\] …………………………(7)
We know that for the quadratic equation \[a{{x}^{2}}+bx+c\] when the discriminant is equal to zero then the quadratic equation has real and equal roots.
From equation (7), we have the value of the discriminant of the quadratic equation \[2{{x}^{2}}-\left( k-1 \right)x+8=0\] . So,
Discriminant = \[{{k}^{2}}-2k-63\] = 0 ………………………(8)
Now, we have to solve the quadratic equation \[{{k}^{2}}-2k-63=0\] and get the value of k.
On factoring the equation (8), we get
\[\begin{align}
  & {{k}^{2}}-2k-63=0 \\
 & \Rightarrow {{k}^{2}}-9k+7k-63=0 \\
 & \Rightarrow k\left( k-9 \right)+7\left( k-9 \right)=0 \\
\end{align}\]
Now, in the above equation, taking \[\left( k-9 \right)\] as common from the whole, we get
\[\Rightarrow \left( k-9 \right)\left( k+7 \right)=0\]
So, \[k=9\] or \[k=-7\] .
Therefore, the value of k for which the equation \[2{{x}^{2}}-\left( k-1 \right)x+8=0\] has real and equal roots is 9 or -7.

Note: We can also solve this question by using the perfect square method.
Since the quadratic equation \[2{{x}^{2}}-\left( k-1 \right)x+8=0\] has equal roots, it must be a perfect square.
Now, transforming the quadratic equation,
\[\Rightarrow 2{{x}^{2}}-\left( k-1 \right)x+8\]
\[\Rightarrow {{\left( \sqrt{2}x \right)}^{2}}-\left( k-1 \right)x+{{\left( 2\sqrt{2} \right)}^{2}}\] ……………………………..(1)
We know the formula, \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\] and \[{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\] .
Since equation (1) is a perfect square, so it can either be of the form \[{{\left( a+b \right)}^{2}}\] or \[{{\left( a-b \right)}^{2}}\] .
Let us assume that equation (1) is of the form \[{{\left( a+b \right)}^{2}}\] .
\[\begin{align}
  & \Rightarrow {{\left( \sqrt{2}x \right)}^{2}}-\left( k-1 \right)x+{{\left( 2\sqrt{2} \right)}^{2}}={{\left( \sqrt{2}x+2\sqrt{2} \right)}^{2}} \\
 & \Rightarrow 2{{x}^{2}}-\left( k-1 \right)x+8=2{{x}^{2}}+8x+8 \\
 & \Rightarrow -\left( k-1 \right)x=8x \\
 & \Rightarrow -\left( k-1 \right)=8 \\
 & \Rightarrow -k+1=8 \\
 & \Rightarrow -k=8-1 \\
 & \Rightarrow -k=7 \\
\end{align}\]
\[\Rightarrow k=-7\] …………………………………(2)
Let us assume that equation (1) is of the form \[{{\left( a-b \right)}^{2}}\] .
\[\begin{align}
  & \Rightarrow {{\left( \sqrt{2}x \right)}^{2}}-\left( k-1 \right)x+{{\left( 2\sqrt{2} \right)}^{2}}={{\left( \sqrt{2}x-2\sqrt{2} \right)}^{2}} \\
 & \Rightarrow 2{{x}^{2}}-\left( k-1 \right)x+8=2{{x}^{2}}-8x+8 \\
 & \Rightarrow -\left( k-1 \right)x=-8x \\
 & \Rightarrow -\left( k-1 \right)=-8 \\
 & \Rightarrow -k+1=-8 \\
\end{align}\]
\[\begin{align}
  & \Rightarrow -k=-8-1 \\
 & \Rightarrow -k=-9 \\
\end{align}\]
\[\Rightarrow k=9\] …………………………………(3)
From equation (2) and equation (3), we have the values of k.
So, \[k=9\] or \[k=-7\] .
Therefore, the value of k for which the equation \[2{{x}^{2}}-\left( k-1 \right)x+8=0\] has real and equal roots is 9 or -7.