Question

answer it ..........​

Answer 1

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Neglect the hand written method on notepad and follow the below method ;

$\blacksquare\:\:\underline{\footnotesize{\text{ Another method of question no. 14}}}$

$\footnotesize{\text{let, the other zeroes are}\:\:\alpha \:\:and \:\:\beta}$

$\footnotesize{\implies(x-1)(x-\alpha)(x-\beta)=0}$

$\footnotesize{\implies(x-1)(x-\alpha)(x-\beta)=0}$

$\footnotesize{\implies(x-1)[x^2+\alpha x+\beta x+\alpha\beta]=0}$

$\footnotesize{\implies x[x^2+\alpha x+\beta x+\alpha\beta]-1[x^2+\alpha x+\beta x+\alpha\beta]=0}$

$\footnotesize{\implies x^3+\alpha x^2+\beta x^2+\alpha\beta x-x^2-\alpha x+\beta x-\alpha\beta=0}$

$\footnotesize{\implies x^3+x^2(\alpha +\beta-1)+x(\alpha\beta-\alpha-\beta)-\alpha\beta=0}$

$\footnotesize{\text{comparing this to the}\:equ^n \:\:x^3+ax^2+bx+c=0\:we\: get}$

$\blacksquare\:\:\footnotesize{\alpha +\beta-1=a}$

$\footnotesize{\alpha +\beta=a+1}$

$\blacksquare\:\:\footnotesize{\alpha\beta-\alpha-\beta=b}$

$\footnotesize{\alpha\beta-(\alpha+\beta)=b}$

$\footnotesize{\alpha\beta-(a+1)=b}$

$\footnotesize{\alpha\beta=b-(a+1)}$

$\footnotesize{\boxed{\alpha\beta=b-a-1}}$

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