Question

For what values of 'k' does the equation $(k-2)x^2+2(2k-3)x+(5k-6)=0$ have equal roots? Find the roots of the equations corresponding to those values of k?

• A. $k=1$, corresponding roots are $-1,-1$
  $k=3$, corresponding roots are $-3,-3$
• B. $k=1$, corresponding roots are $-1,2$
  $k=2$, corresponding roots are $-3,-3$
• C. $k=1$, corresponding roots are $-1,1$
  $k=5$, corresponding roots are $-3,3$
• D. $k=5$, corresponding roots are $-1,1$
  $k=3$, corresponding roots are $-3,-3$

Answer 1

Answer: (A)

$k=1$, corresponding roots are $-1,-1$

$k=3$, corresponding roots are $-3,-3$

$(k-2)x^2+2(2k-3)x+(5k-6)=0$ has equal

roots $\Rightarrow D=0$

$\Rightarrow \left[2\right(2k-3){]}^2-4\times (k-2\left)5\right(k-6)=0$

$\Rightarrow 4k^2-12k+9-5k^2+16k-12=0$

$\Rightarrow -k^2+4k-3=0\Rightarrow (k-1)(k-3)=0$

$\Rightarrow k=1,3$

for $k=1$ eq${}^n$ is $-(x^2+2x+1)=0$

$\Rightarrow (x+1)(x+1)=0$

$\Rightarrow x=-1,-1$

for $k=3$ eq is $x^2+6x+9=0$

$\Rightarrow (x+3{)}^2=0$

$\Rightarrow x=3,-3$