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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
CBSE | Class 11 | Exercise 14.2 | Maths | Quadratic Equations | RD Sharma
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

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

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

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

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

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

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

x=3 \sqrt{2} or x=2 i

\therefore The roots of the given equation are 3 \sqrt{2}, 2 \mathrm{i}

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

apply discriminant rule,

x=\left(-b \pm \sqrt{\left.\left(b^{2}-4 a c\right)\right) / 2 a}\right.

a=1, b=-(5-i), c=(18+i)

    \[\begin{aligned} &x=\frac{-(-(5-i)) \pm \sqrt{(-(5-i))^{2}-4(1)(18+i)}}{2(1)} \\ &=\frac{(5-i) \pm \sqrt{(5-i)^{2}-4(18+i)}}{2} \end{aligned}\]

8 + 14i = 49 – 1 + 14i

So,

48 + 14i = 49 + i2 + 14i [∵ i2 = –1]

= 72 + i2 + 2(7)(i)

= (7 + i)2 [Since, (a + b)2 = a2 + b2 + 2ab]

By using the result 48 + 14i = (7 + i) 2, we get

RD Sharma Solutions for Class 11 Maths Chapter 14 – Quadratic Equations image - 2

x = 2 + 3i or 3 – 4i

∴ The roots of the given equation are 3 – 4i, 2 + 3i

 

 

 

 

 

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