x² - 6x + 9 = 0
Let y = X² - 6x + 9
Step 1:

Step 2:
Points to be plotted: (-4,49), (-3, 36), (-2, 25), (-1, 16), (0, 9), (1, 4), (2, 1), (3, 0), (4, 1)
Step 3:
Draw the parabola and mark the co-ordinates of the intersecting points

Step 4:
The point of intersection of the parabola with x axis is (3, 0)
Since there is only one point of intersection with the x-axis, the quadratic equation has real and equal roots.
∴ Solution (3, 3)

x² + x + 7 = 0
Let y=x2+x+7
Step 1:

Step 2:
Points to be plotted: (-4, 19), (-3, 13), (-2, 9), (-1, 7), (0, 7), (1, 9), (2, 13), (3, 19), (4, 27)
Step 3:
Draw the parabola and mark the co-ordinates of the parabola which intersect with the x-axis.

Step 4:
The roots of the equation are the points of intersection of the parabola with the x axis. Here the parabola does not intersect the x axis at any point.
So, we conclude that there is no real roots for the given quadratic equation.

The distance between the point from the centre is 10 cm.

Length of the tangents PA - PB = 8.7 cm
Construction:
Steps:
(1) With O as centre, draw a circle of radius 5cm.
(2) Draw a line OP = 10 cm.
(3) Draw a perpendicular bisector of OP which cuts OP at M.
(4) With M as centre and MO as radius, draw a circle which cuts previous circle at A and B.
(5) Join AP and BP. AP and BP are the required tangents. Thus length of the tangents are PA and PB = 8.7 cm

Construction:
Step (1) Draw a line segment PQ = 4.5 cm
Step (2) At P, draw PE such that \(\angle QPE={ 35 }^{ 0 }\)
Step (3) At P, draw PF such that \(\angle EPF={ 90 }^{ 0 }\)
Step (4) Draw \(\bot \) bisector to PQ which intersects PF at O.
Step (5) With O centre OP as radius draw a circle.
Step (6) From G, marked arcs of radius 6 cm on the circle marked them as R and S.
Step (7) Joined PR and RQ. Then \(\triangle\)PQR is the required triangle
Step (8) \(\triangle\)PQS is the required triangle

We observe that A is a 2 x 3 matrix and B is a 3 x 3 matrix, hence AB is defined and it will be of the order 2 × 3..
Given A = \({ \left[ \begin{matrix} 1 & 2 & 0 \\ 3 & 1 & 5 \end{matrix} \right] }_{ 2\times 3 }\), B = \({ \left[ \begin{matrix} 8 & 3 & 1 \\ 2 & 4 & 1 \\ 5 & 3 & 1 \end{matrix} \right] }_{ 2x3 }\)
AB = \({ \left[ \begin{matrix} 1 & 2 & 0 \\ 3 & 1 & 5 \end{matrix} \right] }\times \left[ \begin{matrix} 8 & 3 & 1 \\ 2 & 4 & 1 \\ 5 & 3 & 1 \end{matrix} \right] \)
= \(\left[ \begin{matrix} 8+4+0 & 3+8+0 & 1+2+0 \\ 24+2+25 & 9+4+15 & 3+1+5 \end{matrix} \right] =\left[ \begin{matrix} 12 & 11 & 3 \\ 51 & 28 & 9 \end{matrix} \right] \)

The given matrices are equal. Thus all corresponding elements are equal.
Therefore, a - b = 1 …(1)
2a + c = 5 …(2)
2a - b = 0 …(3)
3c + d = 2 …(4)
(3) gives 2a - b = 0
2a = b …(5)
Put 2a = b in equation (1), a - 2a = 1 gives a = −1
Put a = −1 in equation (5), 2(-1) = b gives b = −2
Put a = −1 in equation (2), 2(-1) + c = 5 gives c = 7
Put c = 7 in equation (4), 3(7) + d = 2 gives d = −19
Therefore, a = −1, b = −2, c = 7, d = −19

The general 3 x 3 matrix is given by A = \(\left( \begin{matrix} { a }_{ 11 } & { a }_{ 12 } & { a }_{ 13 } \\ { a }_{ 21 } & { a }_{ 22 } & { a }_{ 23 } \\ { a }_{ 31 } & { a }_{ 32 } & { a }_{ 33 } \end{matrix} \right) \)  a_ij = i²j²
a₁₁ = 1² x 1² = 1 x 1 = 1; a₁₂ = 1² x 2² = 1 x 4 = 4; a₁₃ = 1² x 3² = 1 x 9 = 9
a₂₁ = 2² x 1² = 2 x 1 = 2; a₂₂ = 2² x 2² = 4 x 4 = 16; a₂₃ = 2² x 3² = 4 x 9 = 36
a₃₁ = 3² x 1² = 3 x 1 = 3; a₃₂ = 3² x 2² = 9 x 4 = 36; a₃₃ = 3² x 3² = 9 x 9 = 81
Hence the required matrix is A = \(\left( \begin{matrix} 1 & 4 & 9 \\ 4 & 16 & 36 \\ 9 & 36 & 81 \end{matrix} \right) \)

We know that a matrix of order m x n has mn elements. Thus to find all possible orders of a matrix with 16 elements, we will find all ordered pairs of natural numbers whose product is 16.
Such ordered pairs are (1, 16), (16, 1), (4,4), (8,2), (2,8)
Hence possible orders are 1 x 16, 16 x 1, 4 x 4, 2 x 8, 8 x 2

1³ + 2³ + 3³ +...K³ = \(\left[\frac{k(k+1)}{2}\right]^{2}=44100=(210)^{2}\)
1 + 2 + 3 +...+ k = \(\frac{k(k+1)}{2}=210\)
1 + 2 + 3 +...+ k = 210

1³ + 2³ + 3³ + ...+ 16³ = \(\left[ \frac { 16\times \left( 16+1 \right) }{ 2 } \right] ^{ 2 }\)= (136)² = 18496

1² + 2² +...+ 19² = \(\frac { 19\times \left( 19+1 \right) \left( 2\times 19+1 \right) }{ 6 } =\frac { 19\times 20\times 39 }{ 6 } =2470\)

1+ 2 + 3 + .. + 50
Using , 1 + 2 + 3 + ...+ n = \(\frac { n\left( n+1 \right) }{ 2 } \)
1 + 2 + 3 + ... + 50 = \(\frac { 50\times \left( 50+1 \right) }{ 2 } \) = 1275 

\(1^{3}+2^{3}+3^{3}+\ldots+n^{3}=\left[\frac{n(n+1)}{2}\right]^{2}\)
\(=\left(1^{3}+2^{3}+3^{3}+\ldots+20^{3}\right)-\left(1^{3}+2^{3}+\ldots+9^{3}\right) \)
\(=\left[\frac{20(20+1)}{2}\right]^{2}-\left[\frac{9 \times(9+1)}{2}\right]^{2} \)
\(=\left[\frac{20 \times 21}{2}\right]^{2}-\left[\frac{9 \times 10}{2}\right]^{2}=(210)^{2}-(45)^{2}\)
= 44100 - 2025 = 47075
10³ + 11³ + 12³ +....+ 20³ = 42075

A = \(\left[ \begin{matrix} 1 & 3 \\ 5 & -1 \end{matrix} \right] \)
B = \(\left[ \begin{matrix} 1 & -1 & 2 \\ 3 & 5 & 2 \end{matrix} \right] \)
C = \(\left[ \begin{matrix} 1 & 3 & 2 \\ -4 & 1 & 3 \end{matrix} \right] \) 
\(B+C=\left[\begin{array}{rrr} 1 & -1 & 2 \\ 3 & 5 & 2 \end{array}\right]+\left[\begin{array}{rrr} 1 & 3 & 2 \\ -4 & 1 & 3 \end{array}\right]\)
\(=\left[\begin{array}{rrr} 2 & 2 & 4 \\ -1 & 6 & 5 \end{array}\right]\)
\(A(B+C)=\left[\begin{array}{cc} 1 & 3 \\ 5 & -1 \end{array}\right]\left[\begin{array}{rrr} 2 & 2 & 4 \\ -1 & 6 & 5 \end{array}\right]\)
\(=\left[\begin{array}{rrr} 2-3 & 2+18 & 4+15 \\ 10+1 & 10-6 & 20-5 \end{array}\right]\)
\(=\left[\begin{array}{rrr} -1 & 20 & 19 \\ 11 & 4 & 15 \end{array}\right]\)
\(A B=\left[\begin{array}{cc} 1 & 3 \\ 5 & -1 \end{array}\right]\left[\begin{array}{rrr} 1 & -1 & 2 \\ 3 & 5 & 2 \end{array}\right]\)
\(=\left[\begin{array}{rrr} 1+9 & -1+15 & 2+6 \\ 5-3 & -5-5 & 10-2 \end{array}\right]\)
\(=\left[\begin{array}{rrr} 10 & 14 & 8 \\ 2 & -10 & 8 \end{array}\right]\)
\(A C=\left[\begin{array}{cc} 1 & 3 \\ 5 & -1 \end{array}\right]\left[\begin{array}{rrr} 1 & 3 & 2 \\ -4 & 1 & 3 \end{array}\right]\)
\(=\left[\begin{array}{rrr} 1-12 & 3+3 & 2+9 \\ 5+4 & 15-1 & 10-3 \end{array}\right]\)
\(=\left[\begin{array}{rrr} -11 & 6 & 11 \\ 9 & 14 & 7 \end{array}\right]\)
\(A B+A C=\left[\begin{array}{ccc} 10 & 14 & 8 \\ 2 & -10 & 8 \end{array}\right]+\left[\begin{array}{ccc} -11 & 6 & 11 \\ 9 & 14 & 7 \end{array}\right]\)
\(=\left[\begin{array}{rrr} -1 & 20 & 19 \\ 11 & 4 & 15 \end{array}\right]\)
From (1), (2)
A (B + C) = AB + AC , Hence verified

(B+C) = \(\left[ \begin{matrix} 2 & 3 & 4 \\ 1 & 9 & 2 \\ -7 & 1 & -1 \end{matrix} \right] \)+\(\left[ \begin{matrix} 8 & 3 & 4 \\ 1 & -2 & 3 \\ 2 & 4 & -1 \end{matrix} \right] \)=\(\left[ \begin{matrix} 10 & 6 & 8 \\ 2 & 7 & 5 \\ -5 & 5 & -2 \end{matrix} \right] \)
\(A+(B+C)=\left[ \begin{matrix} 4 & 3 & 1 \\ 2 & 3 & -8 \\ 1 & 0 & -4 \end{matrix} \right] +\left[ \begin{matrix} 10 & 6 & 8 \\ 2 & 7 & 5 \\ -5 & 5 & -2 \end{matrix} \right] =\left[ \begin{matrix} 14 & 9 & 9 \\ 4 & 10 & -3 \\ -4 & 5 & -6 \end{matrix} \right] .... (1)\)
R.H.S = (A+B)+C
\((A+B)=\left[ \begin{matrix} 4 & 3 & 1 \\ 2 & 3 & -8 \\ 1 & 0 & -4 \end{matrix} \right] +\left[ \begin{matrix} 2 & 3 & 4 \\ 1 & 9 & 2 \\ -7 & 1 & -1 \end{matrix} \right] =\left[ \begin{matrix} 6 & 6 & 5 \\ 3 & 12 & -6 \\ -6 & 1 & -5 \end{matrix} \right] \quad \quad \quad (1)\)
\((A+B)+C=\left[ \begin{matrix} 6 & 6 & 5 \\ 3 & 12 & -6 \\ -6 & 1 & -5 \end{matrix} \right] +\left[ \begin{matrix} 8 & 3 & 4 \\ 1 & -2 & 3 \\ 2 & 4 & -1 \end{matrix} \right] =\left[ \begin{matrix} 14 & 9 & 9 \\ 4 & 10 & -3 \\ -4 & 5 & -6 \end{matrix} \right] ...(2)\)
(1) = (2) ⇒ L.H.S. = R.H.S Hence verified.

1² + 2² + 3² +...+ n² = \(\frac{n(n+1)(2 n+1)}{6}\)
With the square colour papers are decorated
= 10² + 11² + 12² + ... + 24²
10² + 11² + 12² + ... + 24² = (1² + 2² + 3² + ...+ 24²) - (1² + 2² + ...+ 9²)
\(=\frac{24 \times(24+1)[2(24)+1]}{6}-\frac{9 \times(9+1)[2(9)+1]}{6} \)
\(=(4 \times 25 \times 49)-\frac{9 \times 10 \times 19}{6} \)
= 4900 - 285 = 4615
4615 cm² area can be decorated