Math, asked by BrainlyHelper, 1 year ago

यदि निम्नलिखित द्विघात समकरणों के मूलों का अस्तित्व हो तो इन्हें पूर्ण वर्ग बनाने की विधि द्वारा ज्ञात कीजिए।
(i) 2x^{2}-7x+3=0
(ii) 2x^{2}+x-4=0
(iii)  4x^{2}+4\sqrt{3}x+3=0
(iv) 2x^{2}+x+4=0

Answers

Answered by abhi178
14
(i) 2x² - 7x + 3 = 0
D = b² - 4ac = (-7)² - 4(3)(2) = 49 - 24 > 0
अतः दिए द्विघात समकरणों के मूलों का अस्तित्व है ।

2x² - 7x + 3 = 0
2x² - 7x = -3
x² - (7/2)x = -3/2
x² - (7/2)x + (7/4)² = -3/2 + (7/4)²
(x - 7/4)² = -3/2 + 49/16 = (-24 + 49)/16
(x - 7/4)2 = (5/4)²
x - 7/4 = ± 5/4
x = (7 ± 5)/4 = 3, 1/2

(ii) 2x² + x - 4 = 0
D = b² - 4ac = 1² - 4(-4)(2) = 33 > 0
अतः दिए द्विघात समकरणों के मूलों का अस्तित्व है ।

2x² + x - 4 = 0
2x² + x = 4
x² + (1/2)x = 2
x² + (1/2)x + (1/4)² = 2 + (1/4)²
x² + 2.(1/4).x + (1/4)² = 2 + 1/16
(x + 1/4)² = 33/16
x + 1/4 = ±√33/4
x = (-1 ± √33)/4

(iii) 4x² + 4√3x + 3 = 0
D = b² - 4ac = (4√3)² - 4(4)(3) = 48 - 48 = 0
अतः दिए द्विघात समकरणों के मूलों का अस्तित्व है ।

4x² + 4√3x + 3 = 0
(2x)² + 2(2x).√3 + (√3)² = 0
(2x + √3)² = 0
x = -√3/2

(iv) 2x² + x + 4 = 0
D = b² - 4ac = 1² - 4(4)(2) < 0
अतः दिए द्विघात समकरणों के मूलों का अस्तित्व नही है ।
Answered by Anonymous
6

Answer:

2x² - 7x + 3 = 0

2x² - 7x = -3

x² - (7/2)x = -3/2

x² - (7/2)x + (7/4)² = -3/2 + (7/4)²

(x - 7/4)² = -3/2 + 49/16 = (-24 + 49)/16

(x - 7/4)2 = (5/4)²

x - 7/4 = ± 5/4

x = (7 ± 5)/4 = 3, 1/2

(ii) 2x² + x - 4 = 0

D = b² - 4ac = 1² - 4(-4)(2) = 33 > 0

अतः दिए द्विघात समकरणों के मूलों का अस्तित्व है ।

2x² + x - 4 = 0

2x² + x = 4

x² + (1/2)x = 2

x² + (1/2)x + (1/4)² = 2 + (1/4)²

x² + 2.(1/4).x + (1/4)² = 2 + 1/16

(x + 1/4)² = 33/16

x + 1/4 = ±√33/4

x = (-1 ± √33)/4

(iii) 4x² + 4√3x + 3 = 0

D = b² - 4ac = (4√3)² - 4(4)(3) = 48 - 48 = 0

अतः दिए द्विघात समकरणों के मूलों का अस्तित्व है ।

4x² + 4√3x + 3 = 0

(2x)² + 2(2x).√3 + (√3)² = 0

(2x + √3)² = 0

x = -√3/2

(iv) 2x² + x + 4 = 0

D = b² - 4ac = 1² - 4(4)(2) < 0

अतः दिए द्विघात समकरणों के मूलों का अस्तित्व नही है ।

hope it helps

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