Find the values of k for each of the following , so that they have two equal roots. (i) 2x2 + kx + 3 = 0 (ii) kx (x – 2) + 6 = 0
Step-by-Step Solution
Key Concept: For a quadratic equation $ax^2+bx+c=0$, the roots are equal (i.e., a repeated root) iff its discriminant $D=b^2-4ac$ is zero.
### (i) $\;2x^{2}+kx+3=0$
1. Identify the coefficients: $a=2$, $b=k$, $c=3$.
2. Write the discriminant: \[D = b^{2}-4ac = k^{2}-4\cdot2\cdot3 = k^{2}-24.\]
3. Set $D=0$ for equal roots: \[k^{2}-24=0 \;\Rightarrow\; k^{2}=24.\]
4. Solve for $k$: \[k = \pm\sqrt{24}=\pm 2\sqrt{6}.\]
### (ii) $\;kx(x-2)+6=0$
1. Expand the expression: \[kx(x-2)+6 = kx^{2}-2kx+6 = 0.\]
2. Identify the coefficients: $a=k$, $b=-2k$, $c=6$.
3. Write the discriminant: \[D = b^{2}-4ac = (-2k)^{2}-4\cdot k \cdot 6 = 4k^{2}-24k.\]
4. Set $D=0$: \[4k^{2}-24k = 0 \;\Rightarrow\; 4k(k-6)=0.\]
5. Solve for $k$: \[k=0 \quad\text{or}\quad k=6.\]
6. Since $k=0$ would eliminate the $x^{2}$ term, the equation would no longer be quadratic (it would reduce to $6=0$, which is impossible). Hence discard $k=0$.
7. Therefore, the required value is $k=6$. For $k=6$, the equation becomes $6x^{2}-12x+6=0 \Rightarrow (x-1)^{2}=0$, confirming equal roots.
Correct Answer: (i) $k = \pm 2\sqrt{6}$
(ii) $k = 6$