Q.1 Find the quadratic polynomial whose sum and product of the zeroes are 21/8 and 5/16 respectively.

 

(OR)

 

Q.2 A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle of 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 𝑚. Find the height of the tree.

We need to find the quadratic polynomial given that the sum and product of the zeroes are 21/8 and 5/16 respectively
It is known that the quadratic equation can be written as
𝑥² − (𝑆𝑢𝑚 𝑜𝑓 𝑟𝑜𝑜𝑡𝑠)𝑥 + (𝑃𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑟𝑜𝑜𝑡𝑠) = 0. For example, if α and β are the roots of a quadratic equation, then the quadratic equation is   2 .
Since it is given that the sum of roots is 21/8  and the product of roots is 5/16 , on substituting
the values, we get the quadratic equation as
𝑥² − (𝑆𝑢𝑚 𝑜𝑓 𝑟𝑜𝑜𝑡𝑠)𝑥 + (𝑃𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑟𝑜𝑜𝑡𝑠) = 0
⇒ 𝑥²− ( 21/8 )𝑥 +   5/16 = 0

⇒   1/16[16𝑥²  −  42𝑥  +  5] =  0
⇒ 16𝑥² − 42𝑥 + 5 = 0
Hence, the required quadratic equation is  16𝑥² − 42𝑥 + 5 = 0

(OR)

We need to find the height of the tree. Let the height of the tree be ℎ.
According to the given data, the diagram can be drawn as shown below
 

Therefore, the height of the tree is ℎ = 𝐵𝐶 + 𝐵𝐷 ... (𝑖)
From ∆𝐵𝐶𝐷,
𝑡𝑎𝑛 30° = 𝐵𝐶 
⇒   1/$\sqrt{3}$    = 𝐵𝐶/CD

= 1/$\sqrt{3}$ = BC/8

BC = 8/$\sqrt{3}$

Similarly,

𝑐𝑜𝑠 30° = 𝐶𝐷/BD 
⇒$\sqrt{3}$/2 =   8/𝐵𝐷
⇒ 𝐵𝐷 = 16/$\sqrt{3}$
On substituting the values, we get the height as
ℎ = 𝐵𝐶 + 𝐵𝐷
⇒ ℎ =   8/$\sqrt{3}$ + 16/$\sqrt{3}$
⇒ ℎ = 24/$\sqrt{3}$

⇒ ℎ = 8$\sqrt{3}$ 𝑚
Hence, the height of the tree is 8$\sqrt{3}$𝑚.