11th NCERT Complex Numbers and Quadratic Equations Exercise 5.3 Questions 10
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Question (1)

Solve the equation x2 + 3 = 0

Solution

x2 + 3 = 0
Comparing to standard form we get, a = 1, b = 0, c = 3
D = b2 - 4 ac = 02 - 4 (1)(3)
= 0 - 12
= -12
= 12i2
√ D = √12 i = 2√3 i
$x = \frac{{ - b \pm \sqrt D }}{{2a}}$ $x = \frac{{ \pm 2\sqrt 3 i}}{2} = \pm \sqrt 3 i$

Question (2)

Solve the equation 2x2 + x + 1 = 0

Solution

2x2 + x + 1 = 0
Comparing to standard form, we get, a = 2, b = 1, and c = 1
$D = {b^2} - 4ac$ $= {\left( 1 \right)^2} - 4\left( 2 \right)\left( 1 \right)$ $= 1 - 8$ $= - 7 = 7{i^2}$ $\sqrt D = \sqrt 7 i$ $x = \frac{{ - b \pm \sqrt D }}{{2a}}$ $= \frac{{ - 1 \pm \sqrt 7 i}}{4}$

Question (3)

Solve the equation x2 + 3x + 9 = 0

Solution

x2 + 3x + 9 = 0
Comparing to standard form, we get, a = 1, b = 3, and c = 9
$D = {b^2} - 4ac$ $= {\left( 3 \right)^2} - 4\left( 1 \right)\left( 9 \right)$ $= 9 - 36$ $= - 27 = 27{i^2}$ $\sqrt D =3 \sqrt 3 i$ $x = \frac{{ - b \pm \sqrt D }}{{2a}}$ $= \frac{{ - 3 \pm 3\sqrt 3 i}}{2}$

Question (4)

Solve the equation –x2 + x – 2 = 0

Solution

- x2 + x - 2 = 0
Comparing to standard form, we get, a = -1, b = 1, and c = -2
$D = {b^2} - 4ac$ $= {\left( 1 \right)^2} - 4\left( -1 \right)\left( -2 \right)$ $= 1 - 8$ $= - 7 = 7{i^2}$ $\sqrt D = \sqrt 7 i$ $x = \frac{{ - b \pm \sqrt D }}{{2a}}$ $= \frac{{ - 1 \pm \sqrt 7 i}}{{ - 2}}$

Question (5)

Solve the equation x2 + 3x + 5 = 0

Solution

x2 + 3x + 5 = 0
Comparing to standard form, we get, a = 1, b = 3, and c = 5
$D = {b^2} - 4ac$ $= {\left( 3 \right)^2} - 4\left( 1 \right)\left( 5 \right)$ $= 9 - 20$ $= - 11 = 11{i^2}$ $\sqrt D = \sqrt 11 i$ $x = \frac{{ - b \pm \sqrt D }}{{2a}}$ $= \frac{{ - 3 \pm \sqrt {11} i}}{2}$

Question (6)

Solve the equation x2 – x + 2 = 0

Solution

x2 - x + 2 = 0
Comparing to standard form, we get, a = 1, b = -1, and c = 2
$D = {b^2} - 4ac$ $= {\left( -1 \right)^2} - 4\left( 1 \right)\left( 2 \right)$ $= 1 - 8$ $= - 7 = 7{i^2}$ $\sqrt D = \sqrt 7 i$ $x = \frac{{ - b \pm \sqrt D }}{{2a}}$ $= \frac{{1 \pm \sqrt 7 i}}{2}$

Question (7)

Solve the equation $\sqrt 2 {x^2} + x + \sqrt 2 = 0$

Solution

$\sqrt 2 {x^2} + x + \sqrt 2 = 0$ Comparing to standard form, we get, a = √2, b = 1, and c = √2
$D = {b^2} - 4ac$ $= {\left( 1 \right)^2} - 4\left( {\sqrt 2 } \right)\left( {\sqrt 2 } \right)$ $= 1 - 8$ $= - 7 = 7{i^2}$ $\sqrt D = \sqrt 7 i$ $x = \frac{{ - b \pm \sqrt D }}{{2a}}$ $= \frac{{ - 1 \pm \sqrt 7 i}}{{2\sqrt 2 }}$

Question (8)

Solve the equation $\sqrt 3 {x^2} - \sqrt 2 x + 3\sqrt 3 = 0$

Solution

$\sqrt 3 {x^2} - \sqrt 2 x + 3\sqrt 3 = 0$ Comparing to standard form, we get, a =√3, b = -√2, and c = 3√3
$D = {b^2} - 4ac$ $= {\left( { - \sqrt 2 } \right)^2} - 4\left( {\sqrt 3 } \right)\left( {3\sqrt 3 } \right)$ $= 2 - 36$ $= - 34 = 34{i^2}$ $\sqrt D = \sqrt 34 i$ $x = \frac{{ - b \pm \sqrt D }}{{2a}}$ $= \frac{{\sqrt 2 \pm \sqrt {34} i}}{{2\sqrt 3 }}$

Question (9)

Solve the equation ${x^2} + x + \frac{1}{{\sqrt 2 }} = 0$

Solution

${x^2} + x + \frac{1}{{\sqrt 2 }} = 0$ $\sqrt 2 {x^2} + \sqrt 2 x + 1 = 0$ Comparing to standard form we get, a = √2, b = √2, c = 1
$D = {b^2} - 4ac$ $= {\left( {\sqrt 2 } \right)^2} - 4\left( {\sqrt 2 } \right)\left( 1 \right)$ $= 2 - 4\sqrt 2$ $= - \left( {4\sqrt 2 - 2} \right) = \left( {4\sqrt 2 - 2} \right){i^2}$ $\sqrt D = \sqrt {4\sqrt 2 - 2} i$ $x = \frac{{ - b \pm \sqrt D }}{{2a}}$ $= \frac{{ - \sqrt 2 \pm \sqrt {4\sqrt 2 - 2} i}}{{2\sqrt 2 }}$ $= \frac{{\sqrt 2 \left[ { - 1 \pm \sqrt {4 - \sqrt 2 } i} \right]}}{{2\sqrt 2 }}$ $= \frac{{ - 1 \pm \sqrt {4 - \sqrt 2 } i}}{2}$

Question (10)

Solve the equation ${x^2} + \frac{x}{{\sqrt 2 }} + 1 = 0$

Solution

${x^2} + \frac{x}{{\sqrt 2 }} + 1 = 0$ $\sqrt 2 {x^2} + x + \sqrt 2 = 0$ Comparing to standard form we get, a = √2 , b = 1 , c = √2
$D = {b^2} - 4ac$ $= {\left( 1 \right)^2} - 4\left( {\sqrt 2 } \right)\left( {\sqrt 2 } \right)$ $= 1 - 8$ $= - 7 = 7{i^2}$ $\sqrt D = \sqrt 7 i$ $x = \frac{{ - b \pm \sqrt D }}{{2a}}$ $= \frac{{ - 1 \pm \sqrt 7 i}}{{2\sqrt 2 }}$