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The equation whose roots are smaller by 1 than those of 2x²-5x+6=0 is.A) 2x² - 9x + 13 = 0 B) 2x² - x + 3 = 0C) 2x² + 9x + 13 = 0D) 2x² + x + 3 = 0I want with Solution plz don't SPAM!​

Answer»

2x^2-5x+6=02x2−5x+6=0

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\textbf{Given:}Given:2x^2-5x+6=02x2−5x+6=0\textbf{To find:}To find:\text{The equation whose roots are smaller by 1 those of $2x^2-5x+6=0$}The equation whose roots are smaller by 1 those of 2x2−5x+6=0\textbf{Solution:}Solution:\text{Let the roots of $2x^2-5x+6=0$ be $\alpha\;\text{and}\;\beta$}Let the roots of 2x2−5x+6=0 be αandβ\text{Then,}Then,\alpha+\beta=\dfrac{-b}{a}=\dfrac{5}{2}α+β=a−b=25\alpha\,\beta=\dfrac{c}{a}=\dfrac{6}{2}=3αβ=ac=26=3\text{We have to find the equation whose roots are $\alpha-1\;\text{and}\;\beta-1$}We have to find the equation whose roots are α−1andβ−1\textbf{Sum of the roots}Sum of the roots(\alpha-1)+(\beta-1)(α−1)+(β−1)=(\alpha+\beta)-2=(α+β)−2=\dfrac{5}{2}-2=25−2=\dfrac{5-4}{2}=25−4=\dfrac{1}{2}=21\textbf{Product of the roots}Product of the roots(\alpha-1)(\beta-1)(α−1)(β−1)=(\alpha\,\beta)-(\alpha+\beta)+1=(αβ)−(α+β)+1=3-\dfrac{5}{2}+1=3−25+1=4-\dfrac{5}{2}=4−25=\dfrac{8-5}{2}=28−5=\dfrac{3}{2}=23\textbf{The quadratic equation is}The quadratic equation isx^2-(\text{Sum of the roots})x+\text{Product of the roots}=0x2−(Sum of the roots)x+Product of the roots=0x^2-\dfrac{1}{2}x+\dfrac{3}{2}=0x2−21x+23=0\implies\bf\,2x^2-x+3=0⟹2x2−x+3=0\therefore\textbf{<klux>OPTION</klux> (b) is correct}∴Option (b) is correct

  • Required ANSWER:-
  • 2x2−x+3=0
  • Hence,Option (B) Is correct✅

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