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  • Help me please, - Complex numbers and quadratic equations - JEE Main-15

 If, for a positive integer n, the quadratic equation,

x(x+1)+(x+1)(x+2)+....

+\left ( x+\overline{n-1} \right )\left ( x+n \right )= 10n

has two consecutive integral solutions, then n is equal to :

  • Option 1)

    9

  • Option 2)

    10

  • Option 3)

    11

  • Option 4)

    12

 

Answers (2)

best_answer

As we learnt in

Condition for Real and distinct roots of Quadratic Equation -

D= b^{2}-4ac> 0

- wherein

ax^{2}+bx+c= 0

is the quadratic equation

 

 

 

x(x+1)+(x+1)(x+2)+...........(x+(\overline{n-1}))(x+n)=10n

(n^{2}+x)+(n^{2}+3x+2)(n^{2}+5x+6)+....................=10n

\Rightarrow\ \; nx^{2}+n^{2}x+\frac{n(n^{2}-31)}{3}=0        \left[\because\ \;1+3+5+7n =^{2} \and \ 2+6+12+.......(n-1)=\frac{n(n-1)(n+1))}{3}\right ]

\Rightarrow\ \; n^{2}+nx+\frac{n^{2}-3}{3}=0

\therefore\ \; n^{2}-\frac{4}{3}(n^{2}-31)\geq0

\therefore\ \; n^{2}\leq124

So maximum value n is11.

Correct option is 3.

 

 


Option 1)

9

This is an incorrect option.

Option 2)

10

This is an incorrect option.

Option 3)

11

This is the correct option.

Option 4)

12

This is an incorrect option.

Posted by

Himanshu

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