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Exercise - 5.3

Complex Numbers And Quadratic Equations

**Question-1 :-** Solve x^{2} + 3 = 0.

We have, x^{2}+ 3 = 0 x^{2}= -3 x = ±√-3 = ±√3i

**Question-2 :-** Solve 2x^{2} + x + 1 = 0.

We have, 2x^{2}+ x + 1 = 0 Now, a = 2; b = 1; c = 1 b^{2}- 4ac = 1^{2}- 4 x 2 x 1 = 1 - 8 = -7 Therefore, the solutions are given by x = (-1 ± √-7)/4 = (-1 ± √7i)/4

**Question-3 :-** Solve x^{2} + 3x + 9 = 0.

We have, x^{2}+ 3x + 9 = 0 Now, a = 1; b = 3; c = 9 b^{2}- 4ac = 3^{2}- 4 x 1 x 9 = 9 - 36 = -27 Therefore, the solutions are given by x = (-3 ± √-27)/2 = (-3 ± 3√3i)/2

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

We have, -x^{2}+ x - 2 = 0 Now, a = -1; b = 1; c = -2 b^{2}- 4ac = 1^{2}- 4 x (-1) x (-2) = 1 - 8 = -7 Therefore, the solutions are given by x = (-1 ± √-7)/(-2) = (-1 ± √7i)/(-2)

**Question-5 :-** Solve x^{2} + 3x + 5 = 0.

We have, x^{2}+ 3x + 5 = 0 Now, a = 1; b = 3; c = 5 b^{2}- 4ac = 3^{2}- 4 x 1 x 5 = 9 - 20 = -11 Therefore, the solutions are given by x = (-3 ± √-11)/2 = (-3 ± √11i)/2

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

We have, x^{2}- x + 2 = 0 Now, a = 1; b = -1; c = 2 b^{2}- 4ac = (-1)^{2}- 4 x 1 x 2 = 1 - 8 = -7 Therefore, the solutions are given by x = (1 ± √-7)/2 = (1 ± √7i)/2

**Question-7 :-** Solve √2x^{2} + x + √2 = 0.

We have, √2x^{2}+ x + √2 = 0 Now, a = √2; b = 1; c = √2 b^{2}- 4ac = 1^{2}- 4 x √2 x √2 = 1 - 8 = -7 Therefore, the solutions are given by x = (-1 ± √-7)/2√2 = (-1 ± √7i)/2√2

**Question-8 :-** Solve √3x^{2} - √2x + 3√3 = 0.

We have, √3x^{2}- √2x + 3√3 = 0 Now, a = √3; b = -√2; c = 3√3 b^{2}- 4ac = (-√2)^{2}- 4 x √3 x 3√3 = 2 - 36 = -34 Therefore, the solutions are given by x = (√2 ± √-34)/2√3 = (√2 ± √34i)/2√3

**Question-9 :-** Solve x^{2} + x + 1/√2 = 0.

We have, x^{2}+ x + 1/√2 = 0 √2x^{2}+ √2x + 1 = 0 Now, a = √2; b = √2; c = 1 b^{2}- 4ac = (√2)^{2}- 4 x √2 x 1 = 2 - 4√2 = -2(2√2 - 1) Therefore, the solutions are given by x = [-√2 ± √-2(2√2 - 1)]/2√2 = [-√2 ± √2√(2√2 - 1)i]/2√2 = [-1 ± √(2√2 - 1)i]/2

**Question-10 :-** Solve x^{2} + x/√2 + 1 = 0.

We have, x^{2}+ x/√2 + 1 = 0 √2x^{2}+ x + √2 = 0 Now, a = √2; b = 1; c = √2 b^{2}- 4ac = 1^{2}- 4 x √2 x √2 = 1 - 8 = -7 Therefore, the solutions are given by x = (-1 ± √-7)/2√2 = (-1 ± √7i)/2√2

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