Question

Form a quadratic polynomial whose one zero is 6 and sum of the zeroes is 0.
4. Find the sum of all natural numbers less than 100 which are divisible by 6.
1
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Answer 1

SOLUTION

TO DETERMINE

1. Form a quadratic polynomial whose one zero is 6 and sum of the zeroes is 0.

2. Find the sum of all natural numbers less than 100 which are divisible by 6.

CONCEPT TO BE IMPLEMENTED

If the Sum of zeroes and Product of the zeroes of a quadratic polynomial is given then the quadratic polynomial is

$\sf{ {x}^{2} -(Sum \: of \: the \: zeroes )x + Product \: of \: the \: zeroes }$

EVALUATION

1. Here it is given that one zero is 6 and sum of the zeroes is 0.

Then other zero = 0 - 6 = - 6

Product of the Zeros

= - 6 × 6

= - 36

Hence the required polynomial

$\sf{ {x}^{2} -(Sum \: of \: the \: zeroes )x + Product \: of \: the \: zeroes }$

$\sf{ = {x}^{2} - 0.x + ( - 36)}$

$\sf{ = {x}^{2} - 36}$

2. We have to find the sum of all natural numbers less than 100 which are divisible by 6.

Now the natural numbers less than 100 which are divisible by 6 are 6 , 12 , 18 , 24, .., 96

This is an arithmetic progression

First term = a = 6

Common Difference = d = 12 - 6 = 6

Let there are n terms

So by the given condition

6 + 6(n-1) = 96

$\implies \sf{6(n - 1) = 90}$

$\implies \sf{(n - 1) = 15}$

$\implies \sf{n = 16}$

Hence the required sum

$\displaystyle \sf{ = \frac{n}{2} \bigg( \: First \: term + Last \: term \bigg) }$

$\displaystyle \sf{ = \frac{16}{2} \bigg( \: 6 + 96\bigg) }$

$\displaystyle \sf{ = \frac{16}{2} \times 102 }$

= 8 × 102

= 816

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