Solve the following quadratic equations:
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) $2 x^{2}-(3+7 i) x+(9 i-3)=0$

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)

$\Rightarrow x^{2}-y^{2}+2 i x y=-2 i$

$\Rightarrow x^{2}-y^{2}=0$ and $2 x y=-2$   …(ii)

Now, $\left(x^{2}+y^{2}\right)^{2}=\left(x^{2}-y^{2}\right)^{2}+4 x^{2} y^{2}$

$\Rightarrow\left(x^{2}+y^{2}\right)^{2}=4$

$\Rightarrow x^{2}+y^{2}=2$  …(iii)

From (ii) and (iii)

$\Rightarrow x=\pm 1$ and $y=\pm 1$

As, $x y$ is negative [from (ii)]

$\Rightarrow x=1, y=-1$ or $, x=-1, y=1$

$\Rightarrow x+i y=1-i$ or,$-1+i$

$\Rightarrow \sqrt{-2 i}=\pm(1-i)$

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, $\left(x^{2}+y^{2}\right)^{2}=\left(x^{2}-y^{2}\right)^{2}+4 x^{2} y^{2}$

$\Rightarrow\left(x^{2}+y^{2}\right)^{2}=49+576=625$

$\Rightarrow x^{2}+y^{2}=25$   …(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) $i x^{2}-4 x-4 i=0$

$\Rightarrow i\left(x^{2}+4 i x-4\right)=0$

$\Rightarrow\left(x^{2}+4 i x-4\right)=0$

$\Rightarrow(x+2 i)^{2}=0$

$\Rightarrow x+2 i=0$

$\Rightarrow x=-2 i$

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, $\left(x^{2}+y^{2}\right)^{2}=\left(x^{2}-y^{2}\right)^{2}+4 x^{2} y^{2}$

$\Rightarrow\left(x^{2}+y^{2}\right)^{2}=225+64=289$

$\Rightarrow x^{2}+y^{2}=17$   …(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 $\frac{1+(4-\sqrt{15}) \mathrm{i}}{4}$ and $\frac{-1-(4+\sqrt{15}) \mathrm{i}}{4}$.

(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, $\left(x^{2}+y^{2}\right)^{2}=\left(x^{2}-y^{2}\right)^{2}+4 x^{2} y^{2}$

$\Rightarrow\left(x^{2}+y^{2}\right)^{2}=9+16=25$

$\Rightarrow x^{2}+y^{2}=5$   ….(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) $i x^{2}-x+12 i=0$

$\Rightarrow i\left(x^{2}+i x+12\right)=0$

$\Rightarrow x^{2}+i x+12=0$

$\Rightarrow x^{2}+4 i x-3 i x+12=0$

$\Rightarrow x(x+4 i)-3 i(x+4 i)=0$

$\Rightarrow(x+4 i)(x-3 i)=0$

$\Rightarrow(x+4 i)=0$ 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, $\left(x^{2}+y^{2}\right)^{2}=\left(x^{2}-y^{2}\right)^{2}+4 x^{2} y^{2}$

$\Rightarrow\left(x^{2}+y^{2}\right)^{2}=196+128=324$

$\Rightarrow x^{2}+y^{2}=18$   …(iii)

From (ii) and (iii)

$\Rightarrow x=\pm 4$ and $y=\pm \sqrt{2}$

As, $x y$ is negative [From (ii)]

$\Rightarrow x=-4, y=\sqrt{2}$ or $, x=4, y=-\sqrt{2}$

$\Rightarrow x+i y=4-\sqrt{2} i$ or,$-4+\sqrt{2} i$

$\Rightarrow \sqrt{14-8 \sqrt{2}} i=\pm(4-\sqrt{2} i)$

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, $\left(x^{2}+y^{2}\right)^{2}=\left(x^{2}-y^{2}\right)^{2}+4 x^{2} y^{2}$

$\Rightarrow\left(x^{2}+y^{2}\right)^{2}=256+900=1156$

$\Rightarrow x^{2}+y^{2}=34$    ….(iii)

From (ii) and (iii)

$\Rightarrow x=\pm 3$ and $y=\pm 5$

As, $x y$ is negative [From (ii)]

$\Rightarrow x=-3, y=5$ or $, x=3, y=-5$

$\Rightarrow x+i y=3-5 i$ or,$-3+5 i$

$\Rightarrow \sqrt{14-8 \sqrt{2}} i=\pm(3-5 i)$

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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