Question
Download Solution PDFजर α आणि β हे द्विघाती बहुपदी f (x) = x2 - 5x +6 चे शून्यक असल्यास, ( α2β + β2α ) चे मूल्य शोधा.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFसंकल्पना:
जर α आणि β ही समीकरण ax2 + bx + c =0 ची मूळे असल्यास,
मुळांची बेरीज (α + β) = \(\rm \frac{-b}{a}\)
मुळांचा गुणाकार (αβ) = \(\rm \frac{c}{a}\)
(x + y)2 = x2 + y2 + 2xy .
गनणा:
दिलेले आहे: f (x) = x2 - 5x + 6
ax2 + bx + c =0 शी f(x) ची तुलना केल्यास, आपल्याकडे आहे, a = 1 , b= -5 आणि c= 6.
आता, मुळांची बेरीज = α + β = \(\rm \frac{-b}{a}\) = \(\rm \frac{-(-5)}{1}\) = 5
आणि मुळांचा गुणाकार αβ = \(\rm \frac{c}{a}\) = \(\rm \frac{6}{1}\) = 6 .
आता, α2β + β2α = αβ ( α+ β )
= 6 × 5
= 30
योग्य पर्याय 2 आहे.
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