Question

Let $f(x)$ and $g(x)$ be two polynomials of degree 2 defined as $f(x)=x^2-2(k-1)x+1$ and $g(x)=ka{x}^2+bx+c$. Given that $a^2+b^2+c^2-2a+6b-4c+14=0,a,b,c\in R$ then
The number of value(s) of $k$ for which $f(x)=0$ and $g(x)=0$ has 3 real and distinct roots provided that their is no common solution among the 2 equation is/are

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (A)

$(a-1{)}^2+(b+3{)}^2+(c-2{)}^2=0$
Hence, $a=1;b=-3;c=2$
$f(x)=x^2-2(k-1)x+1$
$g(x)=k{x}^2-3x+2$
Let $D_1$ be discriminant of $f(x)=0$ and $D_2$ be discriminant of $g(x)=0$
$D_1=4(k-1{)}^2-4;D_2=(9-8k)$
Now,
Case-I : $D_1>0$ and $D_2=0$
$(k-1{)}^2-1>0\text{ and }k=\frac{9}{8}$
Hence, no value of $k$ from this case.
Case-II : $D_2>0$ and $D_1=0$
$9-8k>0$ and $(k-1{)}^2-1=0$
From this case $k=0$ which is rejected.
Hence, no value of $k$ from this case.
Therefore no value of k is nossible.