Question
Download Solution PDFयदि α और β, α > β, बहुपद p(x) = 2x² - 5x + k के शून्यक इस प्रकार हैं कि \(\rm \alpha-\beta=-\frac{7}{2}\) है, तो k निम्नलिखित में से किस बहुपद का शून्यक है?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFदिया गया है:
बहुपद: p(x) = 2x² - 5x + k
शून्यक: α और β, α > β
α - β = -7/2
प्रयुक्त सूत्र:
शून्यकों का योग: α + β = -b/a
शून्यकों का गुणनफल: αβ = c/a
गणना:
शून्यकों का योग: α + β = 5/2
दिया गया है कि, α - β = -7/2
⇒ (α + β) + (α - β) = 5/2 - 7/2
⇒ 2α = -1 ⇒ α = -1/2
⇒ β = 5/2 - (-1/2) ⇒ β = 3
शून्यकों का गुणनफल: αβ = -1/2 × 3
⇒ -3/2 = k/2
⇒ k = -3
k बहुपद x² + 2x - 3 का एक शून्यक है।
इसलिए, सही उत्तर विकल्प (3) है।
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