Let {tex}\alpha{/tex} = 3, {tex}\beta{/tex} = {tex}\frac{1}{2}{/tex} and {tex}\gamma{/tex} = -1. Then,
{tex}( \alpha + \beta + \gamma ) = \left( 3 + \frac { 1 } { 2 } - 1 \right) = \frac { 5 } { 2 }{/tex},
{tex}( \alpha \beta + \beta \gamma + \gamma \alpha ) = \left( \frac { 3 } { 2 } - \frac { 1 } { 2 } - 3 \right) = \frac { - 4 } { 2 }{/tex} = -2
and {tex}\alpha \beta y = \left\{ 3 \times \frac { 1 } { 2 } \times ( - 1 ) \right\} = \frac { - 3 } { 2 }{/tex}
The polynomial  with zeros α,β and {tex}\gamma{/tex} is:
{tex}\mathrm x^3-(\mathrm\alpha+\mathrm\beta+\mathrm\gamma)\mathrm x^2+(\mathrm{αβ}+\mathrm{βγ}+\mathrm{γα})\mathrm x-\mathrm{αβγ}{/tex}

{tex}=\mathrm x^3-\frac52\mathrm x^2-2\mathrm x+\frac32{/tex}
Thus, 2x³- 5x²- 4x + 3 is the desired polynomial.

Given, {tex}\frac{2x}{x-3}+\frac{1}{2x+3}+\frac{3x+9}{(x-3)(2x+3)}=0{/tex}

{tex}\Rightarrow\frac{{2x(2x + 3) + x - 3 + 3x + 9}}{{(x - 3)(2x + 3)}} = 0{/tex}

{tex} \Rightarrow \frac{{4{x^2} + 6x + x - 3 + 3x + 9}}{{2{x^2} + 3x - 6x - 9}} = 0{/tex}

{tex} \Rightarrow \frac{4{x^2} + 10x +6 }{{2{x^2} - 3x - 9}} = 0{/tex}

Cross multiplying equation , we get ,

{tex} \Rightarrow 4{x^2} + 10x +6 = 0{/tex}

{tex} \Rightarrow 4{x^2} + 6x+4x +6 = 0{/tex}

{tex} \Rightarrow x(4{x} + 6)+1(4x +6) = 0{/tex}

{tex}\Rightarrow(x+1)=0{/tex} or {tex} (4x+6)=0{/tex}

{tex}\Rightarrow x=-1{/tex}

{tex}\Rightarrow x=-\frac{3}{2}{/tex}

Hence , the roots of the given quadratic equation are -1 and {tex}-\frac{3}{2}{/tex}

Surface area to colour = surface area of hemisphere + curved surface area of cone

Diameter of hemisphere = 3.5 cm

So radius of hemispherical portion of the lattu = r =  {tex}\frac { 3.5 } { 2 } \mathrm { cm }{/tex} = 1.75 
r = Radius of the concial portion = {tex}\frac{3.5}2{/tex} = 1.75 
 Height of the conical  portion = height of top - radius of hemisphere = {tex}{/tex} 5 - 1.75  = 3.25 cm
Let I be the slant height of the conical part. Then,
{tex}l^2=h^2+r^2{/tex}

{tex}\begin{array}{l}l^2=(3.25)^2+(1.75)^2\\\Rightarrow l^2\;=10.5625+3.0625\\\Rightarrow l^2=13.625\\\Rightarrow l=\sqrt{13.625}\\\Rightarrow l=3.69\end{array}{/tex}
Let S be the total surface area of the top. Then,
{tex}S = 2 \pi r ^ { 2 } + \pi r l{/tex}
{tex}\Rightarrow \quad S = \pi r ( 2 r + l ){/tex}
{tex}\begin{array}{l}\Rightarrow S=\frac{22}7\times1.75(2\times1.75+3.7)\\\;\;\;\;=\;5.5(3.5+3.7)\\=5.5(7.2)\\=39.6\;cm^2\end{array}{/tex}

 
To construct: A line segment of length 8 cm and taking A as centre, to draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Also, to construct tangents to each circle from the centre to the other circle.
Steps of Construction :

1. Bisect BA. Let M be the mid-point of BA.
2. Taking M as centre and MA as radius, draw a circle. Let it intersects the given circle at the points P and Q.
3. Join BP and BQ.
   Then, BP and BQ are the required two tangents from B to the circle with centre A.
4. Again, Let M be the mid-point of AB.
5. Taking M as centre and MB as radius, draw a circle. Let it intersects the given circle at the points R and S.
6. Join AR and AS.
   Then, AR and AS are the required two tangents from A to the circle with centre B.
   Justification: Join BP and BQ.
   Then {tex}\angle APB{/tex} being an angle in the semicircle is 90^o.
   {tex} \Rightarrow BP \bot AP{/tex}
   Since AP is a radius of the circle with centre A, BP has to be a tangent to a circle with centre A. Similarly, BQ is also a tangent to the circle with centre A.
   Again join AR and AS.
   Then {tex}\angle ARB{/tex} being an angle in the semicircle is 90^o.
   {tex} \Rightarrow AR \bot BR{/tex}
   Since BR is a radius of the circle with centre B, AR has to be a tangent to a circle with centre B. Similarly, AS is also a tangent to the circle with centre B.

Given: In the figure, ABC and AMP are two right triangles, right angled at B and M respectively,

To prove:

1. {tex}\triangle {/tex}ABC {tex} \sim {/tex} {tex}\triangle {/tex}AMP
2. {tex}\frac{{CA}}{{PA}} = \frac{{BC}}{{MP}}{/tex}

Proof:

1. In {tex}\triangle {/tex}ABC {tex} \sim {/tex} {tex}\triangle {/tex}AMP
   {tex}\angle{/tex}ABC = {tex}\angle{/tex}AMP (1) ........ [Each equal to 90^o]
   {tex}\angle{/tex}BAC={tex}\angle{/tex}MAP (2).........[Common angle]
   In view of (1) and (2)
   {tex}\triangle {/tex}ABC {tex} \sim {/tex} {tex}\triangle {/tex}AMP ..........AA similarity criterion
2. {tex}\triangle {/tex}ABC {tex} \sim {/tex} {tex}\triangle {/tex}AMP.........Proved above in(i)
   {tex}\therefore {/tex} {tex}\frac{{CA}}{{PA}} = \frac{{BC}}{{MP}}{/tex}.........Corresponding sides of two similar triangles are proportional.

Since, x² = r²sin²Acos²B
y ² = r²sin²A sin²B
and z² = r²cos²A
L.H.S. = x² + y² + z²
{tex}=r^2sin^2Acos^2B+r^2sin^2Asin^2B+r^2cos^2A{/tex}
= {tex}r^2sin^2A(cos^2B+sin^2B)+r^2cos^2A{/tex}
= r²sin² A + r²cos²A
= r²(sin² A + cos²A)
= r².
= R.H.S
Hence Proved.

Let the sum of k terms of A.P. is S_n = 3k² - k
Now k^th term of A.P = S_n - S_n-1
a_k = (3k² - k) - [3 (k - 1)²- ( k - 1)]
= (3k² - k) - [3 (k²- 2k + 1) - ( k - 1)]
= 3k² - k - [ 3k² -6k + 3 - k + 1]
= 3k² - k - [ 3k² -7k + 4 ]
= 3k² - k - 3k² + 7k- 4
= 6k - 4
first term = a = {tex}6 \times 1 - 4 = 6 - 4 = 2{/tex}

We have,
cosec{tex}\theta{/tex} - sin{tex}\theta{/tex} = m and sec{tex}\theta{/tex} - cos{tex}\theta{/tex} = n
{tex}\Rightarrow \quad \frac { 1 } { \sin \theta } - \sin \theta = m \text { and } \frac { 1 } { \cos \theta } - \cos \theta{/tex} = n
{tex}\Rightarrow \quad \frac { 1 - \sin ^ { 2 } \theta } { \sin \theta } = m \text { and } \frac { 1 - \cos ^ { 2 } \theta } { \cos \theta }{/tex} = n
{tex}\Rightarrow \quad \frac { \cos ^ { 2 } \theta } { \sin \theta } = m \text { and } \frac { \sin ^ { 2 } \theta } { \cos \theta }{/tex} = n
{tex}\therefore \quad \left( m ^ { 2 } n \right) ^ { 2 / 3 } + \left( m n ^ { 2 } \right) ^ { 2 / 3 } = \left( \frac { \cos ^ { 4 } \theta } { \sin ^ { 2 } \theta } \times \frac { \sin ^ { 2 } \theta } { \cos \theta } \right) ^ { 2 / 3 } + \left( \frac { \cos ^ { 2 } \theta } { \sin \theta } \times \frac { \sin ^ { 4 } \theta } { \cos ^ { 2 } \theta } \right) ^ { 2 / 3 }{/tex}
= (cos³{tex}\theta{/tex})^2/3 + (sin³{tex}\theta{/tex})^2/3 = cos²{tex}\theta{/tex} + sin²{tex}\theta{/tex} = 1
Hence, (m²n)^2/3 + (mn²)^2/3 = 1

We have,
{tex}8 x ^ { 2 } - 22 x - 21 = 0{/tex}
{tex}\Rightarrow{/tex} 8x² - 28x + 6x - 21 = 0
{tex}\Rightarrow{/tex} 4x(2x - 7) + 3 (2x - 7) = 0
{tex}\Rightarrow{/tex} (2x - 7) (4x + 3) = 0 {tex}\Rightarrow{/tex} 2x - 7 = 0 or, 4x + 3 = 0 {tex}\Rightarrow{/tex} {tex}x = \frac { 7 } { 2 } \text { or } x = - \frac { 3 } { 4 }{/tex}
Thus, {tex}x = \frac { 7 } { 2 } \text { and } x = - \frac { 3 } { 4 }{/tex} are two roots of the equation 8x² - 22x - 21 = 0