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16th KVJ Solutions3-4

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3Q) If squares of the roots of \(x^{4}+bx^{2}+cx+d=0\) are \(\alpha ,\beta ,\gamma ,\delta\), then prove that:

\( 64 \alpha \beta \gamma \delta -[4\sum\alpha\beta -(\sum\alpha)^{2}]^{2} = 0 \).



Let y be the root of  transformed equation.


\(x^{4}+bx^{2}+d=-cx\) since bring all even powers one side and odd powers other side.

Squaring on both sides, we get


\((y^{2}+by+d)^{2}=c^{2}y\) since  \(y=x^{2}\)

\(y^{4}+b^{2}y^{2}+d^{2}+2by^{3}+2bdy+2dy^{2} =c^{2}y\)


It is the equation whose roots are squares of roots of \(x^{4}+bx^{2}+cx+d=0\)

now for (2)\( \alpha, \beta, \gamma, \delta \) are the roots.

\( \sum \alpha\) =-2b

\( \sum \alpha \beta\)=\(b^{2} +2d\)

\( \sum \alpha \beta \gamma\) =-(2bd-c^2)

\( \alpha \beta \gamma \delta \) = \(d^{2}\)

Now plug the values in required function.

\( 64 \alpha \beta \gamma \delta -[4\sum\alpha\beta -(\sum\alpha)^{2}]^{2}\) = \( 64d^{2} -(4(b^{2}+2d)- (-2b)^{2})^{2}\)

=\(64d^{2} - (4b^{2} +8d  -4b^{2})^{2}\)


=\( 64d^{2}- 64d^{2}\)

= 0

Hence it is proved.

16th KVJ Solutions3-4

4Q) a) Let a, b,c be the length of the sides of a triangle and r be the in-radius. Show that:

         r< \(\frac{a^{2}+b^{2}+c^{2}}{3(a+b+c)}\).


 To find this solution , click on the link NEXT on the bottom.


b) A family consists of a grand father, 6 sons & daughters and 4 grand children. They are to be seated in a row for a dinner. The grand children wish to occupy the two seats at each end and the grand father refuses to have a grand child on either side of him. Determine the number of ways in which they can be seated for the dinner.

Solution b) :

  Family consists of

  Grand father  =1

  Sons and daughters = 6

  Grand children  =4

  Total =11 members

  - - - - - - - - - - -   11 places to arrange them in row.

Grand children wish to occupy end seats .

First two and last two seats are occupied by grand children.

They can be arranged in 4x3x2x1 ways =24ways.

grand father refuses to sit behind grand children means he has to be seated 5 places in between as he cannot sit in first 3 and last 3 places.

He can be seated in 5 ways.

Remaining Sons and daughters(6) can be arranged in 6! ways=720 ways

Total number of arrangements =24 x5x720 =86400 ways.

Therefore all members can be seated for dinner in 86,400 ways.

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