In the following, determine the set values of k for
Question:

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) $2 x^{2}+k x-4=0$

Solution:

(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 $k \in R$