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
\[D = {b^2} - 4ac\] \[ = {\left( 1 \right)^2} - 4\left( {\sqrt 2 } \right)\left( {\sqrt 2 } \right)\] \[ = 1 - 8\] \[ = - 7 = 7{i^2}\] \[\sqrt D = \sqrt 7 i\] \[x = \frac{{ - b \pm \sqrt D }}{{2a}}\] \[ = \frac{{ - 1 \pm \sqrt 7 i}}{{2\sqrt 2 }}\]