Find the values of k for which the given quadratic equation has real and distinct roots:

Solution:

(i) The given quadric equation is $k x^{2}+2 x+1=0$, and roots are real and distinct

Then find the value of $k$.

Here,

$a=k, b=2$ and,$c=1$

As we know that $D=b^{2}-4 a c$

Putting the value of $a=k, b=2$ and, $c=1$

$D=(2)^{2}-4 \times k \times 1$

$=4-4 k$

The given equation will have real and distinct roots, if $D>0$

$4-4 k>0$

Now factorizing of the above equation

$4-4 k>0$

$4 k<4$

$k<\frac{4}{4}$

$<1$

Now according to question, the value of k less than 1

Therefore, the value of $k<1$

(ii) The given quadric equation is $k x^{2}+6 x+1=0$, and roots are real and distinct.

Then find the value of k.

Here,

$a=k, b=6$ and,$c=1$

As we know that $D=b^{2}-4 a c$

Putting the value of $a=k, b=6$ and, $c=1$

$D=(6)^{2}-4 \times k \times 1$

$=36-4 k$

The given equation will have real and distinct roots, if $D>0$

$36-4 k>0$

Now factorizing of the above equation

$36-4 k>0$

$4 k<36$

$k<\frac{36}{4}$

$<9$

Now according to question, the value of k less than 9

Therefore, the value of $k<9$

(iii) The given quadric equation is $x^{2}-k x+9=0$, and roots are real and distinct

Then find the value of $k$.

 

Here,

$a=1, b=k$ and,$c=9$

As we know that $D=b^{2}-4 a c$

Putting the value of $a=1, b=k$ and, $c=9$

$D=(k)^{2}-4 \times 1 \times 9$

$=k^{2}-36$

The given equation will have real and distinct roots, if $D>0$

$k^{2}-36>0$

Now factorizing of the above equation

$k^{2}-36>0$

$k^{2}>36$

$k>\sqrt{36}=\pm 6$

$k<-6$ or $k>6$

Therefore, the value of $k<-6$ or, $k>6$