Question

Find the values of k for the quadratic equation $ 2{{x}^{2}}+kx+3=0 $ , so that they have two equal roots.
                                                                                 OR
Find the value of k, for which one root of the quadratic equation $ k{{x}^{2}}-14x+8=0 $ is 2.

Answer 1

Hint: For the first part of the question, we need to find the value of k so that the given equation has equal roots. For this, we will find the value of the discriminant and we will keep it equal to 0. The discriminant will be in terms of k and hence we will get an equation in k. Thus, solving that equation, we will get the required value of k. Then for the or part of the question, we will keep the value of x as 2 in the given equation as it is the root of the equation and hence we will have an equation in terms of k. Thus, we will get the required value of k.

Complete step by step answer:
In the first question, we have been given the equation $ 2{{x}^{2}}+kx+3=0 $ and we have to find the value of k such that this equation has equal roots.
Now, we know that for any quadratic equation to have equal roots, the value of its discriminant ‘D’ must be 0.
Now, we know that D for a quadratic equation $ a{{x}^{2}}+bx+c=0 $ is given as $ D={{b}^{2}}-4ac $
Here, the given equation is $ 2{{x}^{2}}+kx+3=0 $ .
Thus, here we have,
a=2
b=k
c=3
Thus, we get the value of D as:
 $ \begin{align}
  & D={{b}^{2}}-4ac \\
 & D={{k}^{2}}-4\left( 2 \right)\left( 3 \right) \\
 & D={{k}^{2}}-24 \\
\end{align} $
Now, as mentioned above, for the equation to have equal roots, D=0.
Thus, we get:
 $ \begin{align}
  & D=0 \\
 & \Rightarrow {{k}^{2}}-24=0 \\
 & \Rightarrow {{k}^{2}}=24 \\
 & \therefore k=\pm 2\sqrt{6} \\
\end{align} $
Thus, for the given equation to have equal roots, the value of k should be $ \pm 2\sqrt{6} $ .
Now, for the OR part of the first question, we have been given the equation $ k{{x}^{2}}-14x+8=0 $ and we have to find the value of k such that one of the roots of the equation is 2.
Now, we know that the root of the equation satisfies it. Hence, if we put x=2 in the given equation, the equation will be satisfied.
Hence, putting x=2 in the given equation we get:
 $ \begin{align}
  & k{{x}^{2}}-14x+8=0 \\
 & \Rightarrow k{{\left( 2 \right)}^{2}}-14\left( 2 \right)+8=0 \\
 & \Rightarrow 4k-28+8=0 \\
 & \Rightarrow 4k=20 \\
 & \Rightarrow k=\dfrac{20}{4} \\
 & \therefore k=5 \\
\end{align} $
Thus, for this equation to have one root as 2, the value of k should be equal to 5.

Note:
 In the first part of the question, we have used the discriminant. We should know that there are different kinds of roots for different ranges of D.
1. If D>0, roots are different and real
2. If D=0, roots are real and equal
3. If D<0, roots are imaginary
Also remember, when we talk about real roots, they can be irrational too and irrational roots will always occur in pairs of conjugates. Similarly, the imaginary roots will be in the form of ‘iota’ and they will also occur in conjugate pairs.