Exercise 5.3 complex no. ncert math solution class 11

11/03/2023 by Team Gmath

Exercise 5.3(Complex Numbers and Quadratic Equations)

Solve each of the following equations:(Exercise 5.3 complex no. ncert math solution class 11)

Question 1: $x^{2}+3=0$ .

Solution : The given quadratic equation is $x^{2}+3=0$ .

$\Rightarrow x^2 = -3$

$\Rightarrow x= \pm\sqrt{-3}$

$\Rightarrow x = \pm\sqrt{3}i \quad[\because \sqrt{-1}=i]$

Question 2: $2 x^{2}+x+1=0$ .

Solution: The given quadratic equation is $2 x^{2}+x+1=0$ .

On comparing the given equation with $a x^{2}+b x+c=0$ ,

we obtain $a=2, b=1$ , and $c=1$

Therefore, the discriminant of the given equation is given by

$D=b^{2}-4 a c$

$=1^{2}-4 \times 2 \times 1=-7$

Therefore, the required solutions are

$x=\frac{-b \pm \sqrt{D}}{2 a}$

$=\frac{-1 \pm \sqrt{-7}}{2 \times 2}$

$=\frac{-1 \pm \sqrt{7} \cdot \sqrt{-1}}{4}$

$=\frac{-1 \pm \sqrt{7} i}{4} \quad[\because \sqrt{-1}=i]$

Question 3: $x^{2}+3 x+9=0$ .

Solution: The given quadratic equation is $x^{2}+3 x+9=0$ .

On comparing the given equation with $a x^{2}+b x+c=0$ ,

we obtain $a=1, b=3$ , and $c=9$

Therefore, the discriminant of the given equation is given by

$D=b^{2}-4 a c$

$=3^{2}-4 \times 1 \times 9=9-36=-27$

Therefore, the required solutions are

$x=\frac{-b \pm \sqrt{D}}{2 a}$

$=\frac{-3 \pm \sqrt{-27}}{2 \times 1}$

$=\frac{-3 \pm 3 \sqrt{3} \cdot \sqrt{-1}}{2}$

$=\frac{-3 \pm 3 \sqrt{3} i}{2} \quad[\because \sqrt{-1}=i]$

Question 4: $-x^{2}+x-2=0$ .

Solution : The given quadratic equation is $-x^{2}+x-2=0$ .

On comparing the given equation with $a x^{2}+b x+c=0$ ,

we obtain $a=-1, b=1$ , and $c=-2$

Therefore, the discriminant of the given equation is given by

$D=b^{2}-4 a c$

$=1^{2}-4 \times(-1) \times(-2)=-7$

Therefore, the required solutions are

$x=\frac{-b \pm \sqrt{D}}{2 a}$

$=\frac{-1 \pm \sqrt{-7}}{2 \times(-1)}$

$=\frac{-1 \pm \sqrt{7} \cdot \sqrt{-1}}{-2}$

$=\frac{1 \pm \sqrt{7} i}{2} \quad[\because \sqrt{-1}=i]$

Question 5: $x^{2}+3 x+5=0$ .

Solution : The given quadratic equation is $x^{2}+3 x+5=0$ .

On comparing the given equation with $a x^{2}+b x+c=0$ ,

we obtain $a=1, b=3$ , and $c=5$

Therefore, the discriminant of the given equation is given by

$D=b^{2}-4 a c$

$=3^{2}-4 \times 1 \times 5=-11$

Therefore, the required solutions are

$x=\frac{-b \pm \sqrt{D}}{2 a}$

$=\frac{-3 \pm \sqrt{-11}}{2 \times 1}$

$=\frac{-3 \pm \sqrt{11} \cdot \sqrt{-1}}{2}$

$=\frac{-3 \pm \sqrt{11} i}{2} \quad[\because \sqrt{-1}=i]$

Question 6: $x^{2}-x+2=0$ .

Solution: The given quadratic equation is $x^{2}-x+2=0$ .

On comparing the given equation with $a x^{2}+b x+c=0$ ,

we obtain $a=1, b=-1$ , and $c=2$

Therefore, the discriminant of the given equation is given by

$D=b^{2}-4 a c$

$=(-1)^{2}-4 \times 1 \times 2=-7$

Therefore, the required solutions are

$x=\frac{-b \pm \sqrt{D}}{2 a}$

$=\frac{-(-1) \pm \sqrt{-7}}{2 \times 1}$

$=\frac{1 \pm \sqrt{7} \cdot \sqrt{-1}}{2}$

$=\frac{1 \pm \sqrt{7} i}{2} \quad[\because \sqrt{-1}=i]$

Question 7: $\sqrt{2} x^{2}+x+\sqrt{2}=0$ .

Solution: The given quadratic equation is $\sqrt{2} x^{2}+x+\sqrt{2}=0$ .

On comparing the given equation with $a x^{2}+b x+c=0$ ,

we obtain $a=\sqrt{2}, b=1$ , and $c=\sqrt{2}$

Therefore, the discriminant of the given equation is given by

$D=b^{2}-4 a c$

$=1^{2}-4 \times \sqrt{2} \times \sqrt{2}=-7$

Therefore, the required solutions are

$x=\frac{-b \pm \sqrt{D}}{2 a}$

$=\frac{-1 \pm \sqrt{-7}}{2 \times \sqrt{2}}$

$=\frac{-1 \pm \sqrt{7} \cdot \sqrt{-1}}{2 \sqrt{2}}$

$=\frac{-1 \pm \sqrt{7} i}{2 \sqrt{2}} \quad[\because \sqrt{-1}=i]$

Question 8: $\sqrt{3} x^{2}-\sqrt{2} x+3 \sqrt{3}=0$ .

Solution: The given quadratic equation is $\sqrt{3} x^{2}-\sqrt{2} x+3 \sqrt{3}=0$ .

On comparing the given equation with $a x^{2}+b x+c=0$ ,

we obtain $a=\sqrt{3}, b=-\sqrt{2}$ , and $c=3 \sqrt{3}$

Therefore, the discriminant of the given equation is given by

$D=b^{2}-4 a c$

$=(-\sqrt{2})^{2}-4 \times \sqrt{3} \times 3 \sqrt{3}=-34$

Therefore, the required solutions are

$x=\frac{-b \pm \sqrt{D}}{2 a}$

$=\frac{-(-\sqrt{2}) \pm \sqrt{-34}}{2 \times \sqrt{3}}$

$=\frac{\sqrt{2} \pm \sqrt{34} \cdot \sqrt{-1}}{2 \sqrt{3}}$

$=\frac{\sqrt{2} \pm \sqrt{34} i}{2 \sqrt{3}} \quad[\because \sqrt{-1}=i]$

Question 9: $x^{2}+x+\frac{1}{\sqrt{2}}=0$

Solution : The given quadratic equation is $x^{2}+x+\frac{1}{\sqrt{2}}=0$

$\Rightarrow \sqrt{2} x^{2}+\sqrt{2} x+1=0$ .

On comparing the given equation with $a x^{2}+b x+c=0$ ,

we obtain $a=\sqrt{2}, b=\sqrt{2}$ , and $c=1$

Therefore, the discriminant of the given equation is given by

$D=b^{2}-4 a c$

$=\sqrt{2}^{2}-4 \times \sqrt{2} \times 1$

$=2-4 \sqrt{2}$

$=-(4 \sqrt{2}-2)=-2(2 \sqrt{2}-1)$

Therefore, the required solutions are

$x=\frac{-b \pm \sqrt{D}}{2 a}$

$=\frac{-\sqrt{2} \pm \sqrt{-2(2 \sqrt{2}-1)}}{2 \times \sqrt{2}}$

$=\frac{-\sqrt{2} \pm \sqrt{2(2 \sqrt{2}-1)} \cdot \sqrt{-1}}{2 \sqrt{2}}$

$=\frac{-1 \pm \sqrt{2 \sqrt{2}-1} i}{2} \quad[\because \sqrt{-1}=i]$

Question 10: $x^{2}+\frac{x}{\sqrt{2}}+1=0$

Solution: The given quadratic equation is $x^{2}+\frac{x}{\sqrt{2}}+1=0$

$\quad \Rightarrow \sqrt{2} x^{2}+x+\sqrt{2}=0$ .

On comparing the given equation with $a x^{2}+b x+c=0$ ,

we obtain $a=\sqrt{2}, b=1$ , and $c=\sqrt{2}$

Therefore, the discriminant of the given equation is given by

$D=b^{2}-4 a c$

$=1^{2}-4 \times \sqrt{2} \times \sqrt{2}$

$=1-8=-7$

Therefore, the required solutions are

$x=\frac{-b \pm \sqrt{D}}{2 a}$

$=\frac{-1 \pm \sqrt{-7}}{2 \times \sqrt{2}}$

$=\frac{-1 \pm \sqrt{7} \cdot \sqrt{-1}}{2 \sqrt{2}}$

$=\frac{-1 \pm \sqrt{7} i}{2 \sqrt{2}} \quad[\because \sqrt{-1}=i]$

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