In the following, determine the set values of k for which the given quadratic equation has real roots:
(i) $2 x^{2}+3 x+k=0$
(ii) $2 x^{2}+k x+3=0$
(iii) $2 x^{2}-5 x-k=0$
(iv) $k x^{2}+6 x+1=0$
(v) $x^{2}-k x+9=0$
(vi) $2 x^{2}+k x+2=0$
(vii) $3 x^{2}+2 x+k=0$
(viii) $4 x^{2}-3 k x+1=0$
(ix) $2 x^{2}+k x-4=0$
(i) The given quadric equation is $2 x^{2}+3 x+k=0$, and roots are real.
Then find the value of k.
Here, $a=2, b=3$ and, $c=k$
As we know that $D=b^{2}-4 a c$
Putting the value of $a=2, b=3$ and, $c=k$
$=(3)^{2}-4 \times 2 \times k$
$=(3)^{2}-4 \times 2 \times k$
$=9-8 k$
The given equation will have real roots, if $D \geq 0$
$9-8 k \geq 0$
$8 k \leq 9$
$k \leq \frac{9}{8}$
Therefore, the value of $k \leq \frac{9}{8}$
(ii) The given quadric equation is $2 x^{2}+k x+3=0$, and roots are real.
Then find the value of $k$.
Here, $a=2, b=k$ and, $c=3$
As we know that $D=b^{2}-4 a c$
Putting the value of $a=2, b=k$ and,$c=3$
$=(k)^{2}-4 \times 2 \times 3$
$=k^{2}-24$
The given equation will have real roots, if $D \geq 0$
$k^{2}-24 \geq 0$
$k^{2} \geq 24$
$k \geq \sqrt{24}$ or $k \leq-\sqrt{24}$
$k \leq-2 \sqrt{6}$ or $k \geq 2 \sqrt{6}$
Therefore, the value of $k \leq-2 \sqrt{6}$ or $k \geq 2 \sqrt{6}$
(iii) The given quadric equation is $2 x^{2}-5 x-k=0$, and roots are real
Then find the value of $k$.
Here, $a=2, b=-5$ and,$c=-k$
As we know that $D=b^{2}-4 a c$
Putting the value of $a=2, b=-5$ and, $c=-k$
$=(-5)^{2}-4 \times 2 \times(-k)$
$=25+8 k$
The given equation will have real roots, if $D \geq 0$
$25+8 k \geq 0$
$8 k \geq-25$
$k \geq-\frac{25}{8}$
Therefore, the value of $k \geq-\frac{25}{8}$
(iv) The given quadric equation is $k x^{2}+6 x+1=0$, and roots are real
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, $c=1$
$=(6)^{2}-4 \times k \times 1$
$=36-4 k$
The given equation will have real roots, if $D \geq 0$
$36-4 k \geq 0$
$4 k \leq 36$
$k \leq \frac{36}{4}$
$k \leq 9$
Therefore, the value of $k \leq 9$
(v) The given quadric equation is $x^{2}-k x+9=0$, and roots are real
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, $c=9$
$=(-k)^{2}-4 \times 1 \times 9$
$=k^{2}-36$
The given equation will have real roots, if $D \geq 0$
$k^{2}-36 \geq 0$
$k^{2} \geq 36$
$k \geq \sqrt{36}$ or $k \leq-\sqrt{36}$
$k \leq-6$ or $k \geq 6$
Therefore, the value of $k \leq-6$ or $k \geq 6$
(vi) The given quadric equation is $2 x^{2}+k x+2=0$, and roots are real.
Then find the value of $k$.
Here, $a=2, b=k$ and,$c=2$
As we know that $D=b^{2}-4 a c$
Putting the value of $a=2, b=k$ and, $c=2$
$=(k)^{2}-4 \times 2 \times 2$
$=k^{2}-16$
The given equation will have real roots, if $D \geq 0$
$k^{2}-16 \geq 0$
$k^{2} \geq 16$
$k \geq \sqrt{16}$ or $k \leq-\sqrt{16}$
$k \leq-4$ or $k \geq 4$
Therefore, the value of $k \leq-4$ or $k \geq 4$
(vii) The given quadric equation is $3 x^{2}+2 x+k=0$, and roots are real.
Then find the value of $k$.
Here, $a=3, b=2$ and,$c=k$
As we know that $D=b^{2}-4 a c$
Putting the value of $a=3, b=2$ and, $c=k$
$=(2)^{2}-4 \times 3 \times k$
$=4-12 k$
The given equation will have real roots, if $D \geq 0$
$4-12 k \geq 0$
$12 k \leq 4$
$k \leq \frac{4}{12}$
$\leq \frac{1}{3}$
Therefore, the value of $k \leq \frac{1}{3}$
(viii) The given quadric equation is $4 x^{2}-3 k x+1=0$, and roots are real.
Then find the value of k.
Here, $a=4, b=-3 k$ and, $c=1$
As we know that $D=b^{2}-4 a c$
Putting the value of $a=4, b=-3 k$ and, $c=1$
$=(-3 k)^{2}-4 \times 4 \times 1$
$=9 k^{2}-16$
The given equation will have real roots, if $D \geq 0$
$9 k^{2}-16 \geq 0$
$9 k^{2} \geq 16$
$k^{2} \geq \frac{16}{9}$
$k \geq \sqrt{\frac{16}{9}}$
$k \leq-\frac{4}{3}$ or $k \geq \frac{4}{3}$
Therefore, the value of $k \leq-\frac{4}{3}$ or $k \geq \frac{4}{3}$
(ix) The given quadric equation is $2 x^{2}+k x-4=0$, and roots are real
Then find the value of k.
Here, $a=2, b=k$ and, $c=-4$
As we know that $D=b^{2}-4 a c$
Putting the value of $a=2, b=k$ and, $c=-4$
$=(k)^{2}-4 \times 2 \times(-4)$
$=k^{2}+32$
The given equation will have real roots, if $D \geq 0$
$k^{2}+32 \geq 0$
Since left hand side is always positive. So $k \in R$
Therefore, the value of $k \in R$