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Manipulation of a linear equation

Solve the fol...

Question

Solve the following quadratic equations:
(i) $x^{2} - (3 \sqrt{2} + 2 i) x + 6 \sqrt{2 i} = 0$
(ii) $x^{2} - (5 - i) x + (18 + i) = 0$
(iii) $(2 + i) x^{2} - (5 - i) x + 2 (1 - i) = 0$
(iv) $x^{2} - (2 + i) x - (1 - 7 i) = 0$
(v) $i x^{2} - 4 x - 4 i = 0$
(vi) $x^{2} + 4 i x - 4 = 0$
(vii) $2 x^{2} + \sqrt{15} i x - i = 0$
(viii) $x^{2} - x + (1 + i) = 0$
(ix) $i x^{2} - x + 12 i = 0$
(x) $x^{2} - (3 \sqrt{2} - 2 i) x - \sqrt{2} i = 0$
(xi) $x^{2} - (\sqrt{2} + i) x + \sqrt{2} i = 0$
(xii) $2 x^{2} - (3 + 7 i) x + (9 i - 3) = 0$

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Solution

$(i) x^{2} - (3 \sqrt{2} + 2 i) x + 6 \sqrt{2} i = 0 \Rightarrow x^{2} - 3 \sqrt{2} x - 2 i x + 6 \sqrt{2} i = 0 \Rightarrow x (x - 3 \sqrt{2}) - 2 i (x - 3 \sqrt{2}) = 0 \Rightarrow (x - 3 \sqrt{2}) (x - 2 i) = 0 \Rightarrow (x - 3 \sqrt{2}) = 0 or (x - 2 i) = 0 \Rightarrow x = 3 \sqrt{2}, 2 i So, the roots of the given quadratic equation are 3 \sqrt{2} and 2 i .$

$(ii) x^{2} - (5 - i) x + (18 + i) = 0 Comparing the given equation with the general form a x^{2} + b x + c = 0, we get a = 1, b = - (5 - i) and c = (18 + i) x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \Rightarrow x = \frac{(5 - i) \pm \sqrt{{(5 - i)}^{2} - 4 (18 + i)}}{2} \Rightarrow x = \frac{(5 - i) \pm \sqrt{{(5 - i)}^{2} - 4 (18 + i)}}{2} \Rightarrow x = \frac{(5 - i) \pm \sqrt{- 48 - 14 i}}{2} \Rightarrow x = \frac{(5 - i) \pm i \sqrt{48 + 14 i}}{2} \Rightarrow x = \frac{(5 - i) \pm i \sqrt{49 - 1 + 2 \times 7 \times i}}{2} \Rightarrow x = \frac{(5 - i) \pm i \sqrt{{(7 + i)}^{2}}}{2} \Rightarrow x = \frac{(5 - i) \pm i (7 + i)}{2} \Rightarrow x = \frac{(5 - i) + i (7 + i)}{2} or x = \frac{(5 - i) - i (7 + i)}{2} \Rightarrow x = 2 + 3 i, 3 - 4 i So, the roots of the given quadratic equation are 2 + 3 i and 3 - 4 i .$

$(iii) (2 + i) x^{2} - (5 - i) x + 2 (1 - i) = 0 Comparing the given equation with the general form a x^{2} + b x + c = 0, we get a = (2 + i), b = - (5 - i) and c = 2 (1 - i) x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \Rightarrow x = \frac{(5 - i) \pm \sqrt{{(5 - i)}^{2} - 4 (2 + i) 2 (1 - i)}}{2 (2 + i)} \Rightarrow x = \frac{(5 - i) \pm \sqrt{- 2 i}}{2 (2 + i)} \Rightarrow x = \frac{(5 - i) \pm \sqrt{- 2 i}}{2 (2 + i)} . . . (i) Let x + i y = \sqrt{- 2 i} . Then, \Rightarrow {(x + i y)}^{2} = - 2 i \Rightarrow x^{2} - y^{2} + 2 i x y = - 2 i \Rightarrow x^{2} - y^{2} = 0 and 2 x y = - 2 . . . (ii) Now, {(x^{2} + y^{2})}^{2} = {(x^{2} - y^{2})}^{2} + 4 x^{2} y^{2} \Rightarrow {(x^{2} + y^{2})}^{2} = 4 \Rightarrow x^{2} + y^{2} = 2 . . . (iii) From (ii) and (iii) \Rightarrow x = \pm 1 a n d y = \pm 1 As, x y is negative [from (ii)] \Rightarrow x = 1, y = - 1 o r, x = - 1, y = 1 \Rightarrow x + i y = 1 - i o r, - 1 + i \Rightarrow \sqrt{- 2 i} = \pm (1 - i) Substituting this value in (i), we get \Rightarrow x = \frac{(5 - i) \pm (1 - i)}{2 (2 + i)} \Rightarrow x = 1 - i, \frac{4}{5} - \frac{2}{5} i So, the roots of the given quadratic equation are 1 - i and \frac{4}{5} - \frac{2}{5} i .$

$(iv) x^{2} - (2 + i) x - (1 - 7 i) = 0 Comparing the given equation with the general form a x^{2} + b x + c = 0, we get a = 1, b = - (2 + i) and c = - (1 - 7 i) x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \Rightarrow x = \frac{(2 + i) \pm \sqrt{{(2 + i)}^{2} + 4 (1 - 7 i)}}{2} \Rightarrow x = \frac{(2 + i) \pm \sqrt{7 - 24 i}}{2} . . . (i) Let x + i y = \sqrt{7 - 24 i} . Then, \Rightarrow {(x + i y)}^{2} = 7 - 24 i \Rightarrow x^{2} - y^{2} + 2 i x y = 7 - 24 i \Rightarrow x^{2} - y^{2} = 7 and 2 x y = - 24 . . . (ii) Now, {(x^{2} + y^{2})}^{2} = {(x^{2} - y^{2})}^{2} + 4 x^{2} y^{2} \Rightarrow {(x^{2} + y^{2})}^{2} = 49 + 576 = 625 \Rightarrow x^{2} + y^{2} = 25 . . . (iii) From (ii) and (iii) \Rightarrow x = \pm 4 and y = \pm 3 As, x y is negative [From (ii)] \Rightarrow x = - 4, y = 3 or, x = 4, y = - 3 \Rightarrow x + i y = - 4 + 3 i or, 4 - 3 i \Rightarrow \sqrt{7 - 24 i} = \pm 4 - 3 i Substituting these values in (i), we get \Rightarrow x = \frac{(2 + i) \pm (4 - 3 i)}{2} \Rightarrow x = 3 - i, - 1 + 2 i So, the roots of the given quadratic equation are 3 - i and - 1 + 2 i .$

$(v) i x^{2} - 4 x - 4 i = 0 \Rightarrow i (x^{2} + 4 i x - 4) = 0 \Rightarrow (x^{2} + 4 i x - 4) = 0 \Rightarrow {(x + 2 i)}^{2} = 0 \Rightarrow x + 2 i = 0 \Rightarrow x = - 2 i So, the roots of the given quadratic equation are - 2 i and - 2 i .$

$(vi) x^{2} + 4 i x - 4 = 0 \Rightarrow x^{2} + 2 \times x \times 2 i + {(2 i)}^{2} = 0 \Rightarrow {(x + 2 i)}^{2} = 0 \Rightarrow x + 2 i = 0 \Rightarrow x = - 2 i So, the roots of the given quadratic equation are - 2 i and - 2 i .$

$(vii) 2 x^{2} + \sqrt{15} i x - i = 0 Comparing the given equation with the general form a x^{2} + b x + c = 0, we get a = 2, b = \sqrt{15} i and c = - i x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \Rightarrow x = \frac{- \sqrt{15} i \pm \sqrt{{(\sqrt{15} i)}^{2} + 8 i}}{4} \Rightarrow x = \frac{- \sqrt{15} i \pm \sqrt{8 i - 15}}{4} . . . (i) Let x + i y = \sqrt{8 i - 15} . Then, \Rightarrow {(x + i y)}^{2} = 8 i - 15 \Rightarrow x^{2} - y^{2} + 2 i x y = 8 i - 15 \Rightarrow x^{2} - y^{2} = - 15 and 2 x y = 8 . . . (ii) Now, {(x^{2} + y^{2})}^{2} = {(x^{2} - y^{2})}^{2} + 4 x^{2} y^{2} \Rightarrow {(x^{2} + y^{2})}^{2} = 225 + 64 = 289 \Rightarrow x^{2} + y^{2} = 17 . . . (iii) From (ii) and (iii) \Rightarrow x = \pm 1 and y = \pm 4 As, x y is positive [From (ii)] \Rightarrow x = 1, y = 4 or, x = - 1, y = - 4 \Rightarrow x + i y = 1 + 4 i or, - 1 - 4 i \Rightarrow \sqrt{8 i - 15} = \pm (1 - 4 i) Substituting these values in (i), we get, \Rightarrow x = \frac{- \sqrt{15} i \pm (1 + 4 i)}{4} \Rightarrow x = \frac{1 + (4 - \sqrt{15}) i}{4}, \frac{- 1 - (4 + \sqrt{15}) i}{4} So, the roots of the given quadratic equation are \frac{1 + (4 - \sqrt{15}) i}{4} and \frac{- 1 - (4 + \sqrt{15}) i}{4} .$

$(viii) x^{2} - x + (1 + i) = 0 Comparing the given equation with the general form a x^{2} + b x + c = 0, we get a = 1, b = - 1 and c = (1 + i) x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \Rightarrow x = \frac{1 \pm \sqrt{1 - 4 (1 + i)}}{2} \Rightarrow x = \frac{1 \pm \sqrt{- 3 - 4 i}}{2} . . . (i) Let x + i y = \sqrt{- 3 - 4 i} . Then, \Rightarrow {(x + i y)}^{2} = - 3 - 4 i \Rightarrow x^{2} - y^{2} + 2 i x y = - 3 - 4 i \Rightarrow x^{2} - y^{2} = - 3 and 2 x y = - 4 . . . (ii) Now, {(x^{2} + y^{2})}^{2} = {(x^{2} - y^{2})}^{2} + 4 x^{2} y^{2} \Rightarrow {(x^{2} + y^{2})}^{2} = 9 + 16 = 25 \Rightarrow x^{2} + y^{2} = 5 . . . (iii) From (ii) and (iii) \Rightarrow x = \pm 1 and y = \pm 2 As, x y is negative [From (ii)] \Rightarrow x = 1, y = - 2 or, x = - 1, y = 2 \Rightarrow x + i y = 1 - 2 i or - 1 + 2 i \Rightarrow \sqrt{- 3 - 4 i} = \pm (1 - 2 i) Substituting these values in (i), we get \Rightarrow x = \frac{1 \pm (1 - 2 i)}{2} \Rightarrow x = 1 - i, i So, the roots of the given quadratic equation are 1 - i and i .$

$(ix) i x^{2} - x + 12 i = 0 \Rightarrow i (x^{2} + i x + 12) = 0 \Rightarrow x^{2} + i x + 12 = 0 \Rightarrow x^{2} + 4 i x - 3 i x + 12 = 0 \Rightarrow x (x + 4 i) - 3 i (x + 4 i) = 0 \Rightarrow (x + 4 i) (x - 3 i) = 0 \Rightarrow (x + 4 i) = 0 or (x - 3 i) = 0 \Rightarrow x = - 4 i, 3 i So, the roots of the given quadratic equation are - 4 i and 3 i .$

$(x) x^{2} - (3 \sqrt{2} - 2 i) x - \sqrt{2} i = 0 Comparing the given equation with the general form a x^{2} + b x + c = 0, we get a = 1, b = - (3 \sqrt{2} - 2 i) and c = - \sqrt{2} i x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \Rightarrow x = \frac{(3 \sqrt{2} - 2 i) \pm \sqrt{{(3 \sqrt{2} - 2 i)}^{2} + 4 \sqrt{2} i}}{2} \Rightarrow x = \frac{(3 \sqrt{2} - 2 i) \pm \sqrt{14 - 8 \sqrt{2} i}}{2} . . . (i) Let x + i y = \sqrt{14 - 8 \sqrt{2} i} . Then, \Rightarrow {(x + i y)}^{2} = 14 - 8 \sqrt{2} i \Rightarrow x^{2} - y^{2} + 2 i x y = 14 - 8 \sqrt{2} i \Rightarrow x^{2} - y^{2} = 14 and 2 x y = - 8 \sqrt{2} . . . (ii) Now, {(x^{2} + y^{2})}^{2} = {(x^{2} - y^{2})}^{2} + 4 x^{2} y^{2} \Rightarrow {(x^{2} + y^{2})}^{2} = 196 + 128 = 324 \Rightarrow x^{2} + y^{2} = 18 . . . (iii) From (ii) and (iii) \Rightarrow x = \pm 4 and y = \pm \sqrt{2} As, x y is negative [From (ii)] \Rightarrow x = - 4, y = \sqrt{2} o r, x = 4, y = - \sqrt{2} \Rightarrow x + i y = 4 - \sqrt{2} i o r, - 4 + \sqrt{2} i \Rightarrow \sqrt{14 - 8 \sqrt{2} i} = \pm (4 - \sqrt{2} i) Substituting these values in (i), we get \Rightarrow x = \frac{(3 \sqrt{2} - 2 i) \pm (4 - \sqrt{2} i)}{2} \Rightarrow x = \frac{(3 \sqrt{2} + 4) - i (2 + \sqrt{2})}{2}, \frac{(3 \sqrt{2} - 4) - i (2 - \sqrt{2})}{2} So, the roots of the given quadratic equation are \frac{(3 \sqrt{2} + 4) - i (2 + \sqrt{2})}{2} and \frac{(3 \sqrt{2} - 4) - i (2 - \sqrt{2})}{2} .$

$(xi) x^{2} - (\sqrt{2} + i) x + \sqrt{2} i = 0 Comparing the given equation with the general form a x^{2} + b x + c = 0, we get a = 1, b = - (\sqrt{2} + i) and c = \sqrt{2} i x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \Rightarrow x = \frac{(\sqrt{2} + i) \pm \sqrt{{(\sqrt{2} + i)}^{2} - 4 \sqrt{2} i}}{2} \Rightarrow x = \frac{(\sqrt{2} + i) \pm \sqrt{1 - 2 \sqrt{2} i}}{2} \Rightarrow x = \frac{(\sqrt{2} + i) \pm \sqrt{{(\sqrt{2})}^{2} - 1^{2} - 2 \sqrt{2} i}}{2} \Rightarrow x = \frac{(\sqrt{2} + i) \pm \sqrt{{(\sqrt{2} - i)}^{2}}}{2} \Rightarrow x = \frac{(\sqrt{2} + i) \pm (\sqrt{2} - i)}{2} \Rightarrow x = \sqrt{2}, i So, the roots of the given quadratic equation are \sqrt{2} and i .$

$(xii) 2 x^{2} - (3 + 7 i) x + (9 i - 3) = 0 Comparing the given equation with the general form a x^{2} + b x + c = 0, we get a = 2, b = - (3 + 7 i) and c = (9 i - 3) x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \Rightarrow x = \frac{(3 + 7 i) \pm \sqrt{{(3 + 7 i)}^{2} - 8 (9 i - 3)}}{4} \Rightarrow x = \frac{(3 + 7 i) \pm \sqrt{- 16 - 30 i}}{4} . . . (i) Let x + i y = \sqrt{- 16 - 30 i} . Then, \Rightarrow {(x + i y)}^{2} = - 16 - 30 i \Rightarrow x^{2} - y^{2} + 2 i x y = - 16 - 30 i \Rightarrow x^{2} - y^{2} = - 16 and 2 x y = - 30 . . . (ii) Now, {(x^{2} + y^{2})}^{2} = {(x^{2} - y^{2})}^{2} + 4 x^{2} y^{2} \Rightarrow {(x^{2} + y^{2})}^{2} = 256 + 900 = 1156 \Rightarrow x^{2} + y^{2} = 34 . . . (iii) From (ii) and (iii) \Rightarrow x = \pm 3 and y = \pm 5 As, x y is negative [From (ii)] \Rightarrow x = - 3, y = 5 o r, x = 3, y = - 5 \Rightarrow x + i y = 3 - 5 i o r, - 3 + 5 i \Rightarrow \sqrt{14 - 8 \sqrt{2} i} = \pm (3 - 5 i) Substituting these values in (i), we get \Rightarrow x = \frac{(3 + 7 i) \pm (3 - 5 i)}{4} \Rightarrow x = \frac{3 + i}{2}, 3 i So, the roots of the given quadratic equation are \frac{3 + i}{2} and 3 i .$

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Similar questions

Solve the following quadratic equations :

(i) $x^{2} - (3 \sqrt{2} + 2 i) x + 6 \sqrt{2} i = 0$

(ii) $x^{2} - (5 - i) x + (18 + i) = 0$

(iii) $(2 + i) x^{2} - (5 - i) x + 2 (1 - i) = 0$

(iv) $x^{2} - (2 + i) x - (1 - 7 i) = 0$

(v) $i x^{2} - 4 x - 4 i = 0$

(vi) $x^{2} + 4 i x - 4 = 0$

(vii) $2 x^{2} + \sqrt{15} i x - i = 0$

(viii) $x^{2} - x + (1 + i) = 0$

(ix) $i x^{2} - x + 12 i = 0$

(x) $x^{2} - (3 \sqrt{2} - 2 i) x - \sqrt{2} i = 0$

(xi) $x^{2} - (\sqrt{2} + i) x + \sqrt{2} i = 0$

(xii) $2 x^{2} - (3 + 7 i) x + (9 i - 3) = 0$

Solve the following quadratic equations by factorization method :

(i) $x^{2} + 10 i x - 21 = 0$

(ii) $x^{2} + (1 - 2 i) x - 2 i = 0$

(iii) $x^{2} - (2 \sqrt{3} + 3 i) x + 6 \sqrt{3} i = 0$

(iv) $6 x^{2} - 17 i x - 12 = 0$

Q. In the following, determine whether the given quadratic equations have real roots and if so, find the roots:

(i) 16x² = 24x + 1
(ii) x² + x + 2 = 0
(iii)

\sqrt{3} x^{2} + 10 x - 8 \sqrt{3} = 0

(iv) 3x² − 2x + 2 = 0
(v)

2 x^{2} - 2 \sqrt{6} x + 3 = 0

3 a^{2} x^{2} + 8 a b x + 4 b^{2} = 0, a \neq 0

3 x^{2} + 2 \sqrt{5} x - 5 = 0

(viii) x² − 2x + 1 = 0
(ix)

2 x^{2} + 5 \sqrt{3} x + 6 = 0

\sqrt{2} x^{2} + 7 x + 5 \sqrt{2} = 0

2 x^{2} - 2 \sqrt{2} x + 1 = 0

(xii) 3x² − 5x + 2 = 0

Q. Find the roots of the quadratic equations by using the quadratic formula in the following equation:
(i)

2 x^{2} - 3 x - 5 = 0

5 x^{2} + 13 x + 8 = 0

- 3 x^{2} + 5 x + 12 = 0

- x^{2} + 7 x - 10 = 0

x^{2} + 2 \sqrt{2} x - 6 = 0

x^{2} - 3 \sqrt{5} x + 10 = 0

\frac{1}{2} x^{2} - \sqrt{11} x + 1 = 0

Q. Solving the following quadratic equations by factorization method:
(i)

x^{2} + 10 i x - 21 = 0

x^{2} + (1 - 2 i) x - 2 i = 0

x^{2} - (2 \sqrt{3} + 3 i) x + 6 \sqrt{3} i = 0

6 x^{2} - 17 i x - 12 = 0

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