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RBSE Solutions for Class 11 Maths Chapter 5 सम्मिश्र संख्याएँ Ex 5.4

June 11, 2019 by Prasanna Leave a Comment

Rajasthan Board RBSE Class 11 Maths Chapter 5 सम्मिश्र संख्याएँ Ex 5.4

प्रश्न 1.
निम्न, समीकरणों के हल वैदिक विधि से ज्ञात कीजिए
(i) x² + 4x + 13 = 0
(ii) 2x² + 5x + 4 = 0
(iii) ix² + 4x – \(\frac { 15 }{ 2 }\) = 0
हल-
(i) यहाँ पर D1 = 2x + 4,
RBSE Solutions for Class 11 Maths Chapter 5 सम्मिश्र संख्याएँ Ex 5.4

(ii) 2x² + 5x + 4 = 0
यहाँ पर D1 = 4x + 5
RBSE Solutions for Class 11 Maths Chapter 5 सम्मिश्र संख्याएँ Ex 5.4

(iii) ix² + 4x – \(\frac { 15 }{ 2 }\) = 0
यहाँ पर D1 = 2ix + 4
RBSE Solutions for Class 11 Maths Chapter 5 सम्मिश्र संख्याएँ Ex 5.4
RBSE Solutions for Class 11 Maths Chapter 5 सम्मिश्र संख्याएँ Ex 5.4
RBSE Solutions for Class 11 Maths Chapter 5 सम्मिश्र संख्याएँ Ex 5.4

प्रश्न 2.
द्विघात समीकरण ज्ञात कीजिए जिसके मूल
(i) 5 तथा -2 हैं।
(ii) 1 + 2i
हल-
(a) दिया है
α = 5, β = -2
अभीष्ट समीकरण
x² – (मूलों का योग)x + मूलों का गुणनफल = 0
x² – (α + β)x + αβ = 0
x² – (5 – 2)x + (5) x (-2) = 0
x² – 3x – 10 = 0

(ii) 1 + 2i
दिये गये मूल का संयुग्मी मूल = 1 – 2i
∴ α = 1 + 2i और β = 1 – 2i
अभीष्ट समीकरण
x² – (मूलों का योग)x + मूलों को गुणनफल = 0
x² – (1 + 2i + 1 – 2i)x + (1 + 2i) × (1 – 2i) = 0
x² – 2x + (1)² – (2i)² = 0
x² – 2x + 1 + 4 = 0
x² – 2x + 5 = 0

प्रश्न 3.
यदि समीकरण x² – px + q = 0 का एक मूल दूसरे का दुगुना है तो सिद्ध कीजिए कि 2p² = 9q.
हल-
माना दिये गये समीकरण x² – px + q = 0 के मूल α तथा β है|
RBSE Solutions for Class 11 Maths Chapter 5 सम्मिश्र संख्याएँ Ex 5.4
RBSE Solutions for Class 11 Maths Chapter 5 सम्मिश्र संख्याएँ Ex 5.4

प्रश्न 4.
वह प्रतिबन्ध ज्ञात कीजिए जिसमें समीकरण ax² + bx + c = 0 के मूल m : n के अनुपात में हैं।
हल-
माना दिये गये समीकरण ax² + bx + c = 0 के मूल αm तथा αn हैं।
∴ मूलों का योग = \(-\frac { b }{ a }\) ∴ αm + αn = \(-\frac { b }{ a }\)
RBSE Solutions for Class 11 Maths Chapter 5 सम्मिश्र संख्याएँ Ex 5.4
⇒ b²mn = ac(m + n)²
⇒ mnb² = ac(m + n)²

RBSE Solutions for Class 11 Maths

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