x²-9=0
Let y=x²-9
Step 1:

Step 2:
The points to be plotted: (-4,7), (-3, 0), (-2, -5), (-1, -8), (0, -9), (1, -8), (2, -5), (3, 0), (4, 7)
(v) Real and equal roots
Step 3: 
Draw the parabola and mark the co-ordinates of the parabola which intersect the x-axis.

Step 4:
The roots of the equation are the co-ordinates of the intersecting points (-3, 0) and (3, 0) of the parabola with the x-axis which are -3 and 3 respectively.
Step 5:
Since there are two points of intersection with the x axis, the quadratic equation has real and unequal roots.
∴ Solution{-3, 3}

x² + 2x + 5 = 0
Let y = x² + 2x + 5
Step 1 Prepare a table of values for the equation y = x² + 2x + 5

Step 2: Plot the above ordered pairs(x, y) on the graph using suitable scale.

Step 3: Join the points by a free-hand smooth curve this smooth curve is the graph of  y = x² + 2x + 5
Step 4: The solutions of the given quadratic equation are the x coordinates of the intersecting points of the parabola the X axis.
Here the parabola doesn’t intersect or touch the X axis.
So, we conclude that there is no real root for the given quadratic equation.

Construction:
Steps (1) Draw QR = 6.5 cm.
Steps (2) Draw \(\angle RQE={ 60 }^{ 0 }\)
Steps (3) Draw \(\angle FQE={ 90 }^{ 0 }\)
Steps (4) Drawn the perpendicular bisector XY to ER which intersects QF at O and ER at G.
Steps (5)  With O as center and OQ as radius drawn a circle
Steps (6) XY intersects QR at G. On XY, from G marked an arc at M, such that GM = 4.5 cm
Steps (7) Drawn AB through M which is parallel to QR
Steps (8)  AB meets the circle at P and S
Steps (9) Joined QP and RP Then \(\triangle\)PQR is the required triangle.
Steps (10) Here \(\triangle\)SQR is also another required triangle.

Total number of cards = 52
Event of selecting a spade card = A
Event of selecting a heart card = B
n(A) = 13, n(B) = 13
P(A) =\(\frac { 13 }{ 52 } =\frac { 1 }{ 4 } \), P(B) =\(\frac { 13 }{ 52 } =\frac { 1 }{ 4 } \).
A ⋂ B refers to a card is both spade and heart. It is not possible

A, B are exclusive events.
P(A,B) = P(A) + P(B) =\(\frac { 1 }{ 4 } +\frac { 1 }{ 4 } =\frac { 1 }{ 2 } \).

σ  = 21.2, \(\bar { x } \)  = 36.6
CV = \(\frac { \sigma }{ \bar { x } } \) x 100 = \(\frac { 21.2 }{ 36.6 } \) x 100 = 57.92%

Let a natural number be x.
The next number = x + 1
x (x + 1) = 20
X²+x-20 = 0
(x + 5)(x - 4) = 0
x=-5,4
∴ x=4
(∵ x≠-5, x is a natural number
The next number = 4 + 1= 5
Two consecutive numbers are 4,5.

f(2) - f(4)
f(2) = 2x - 1
= 2(2) - 1 = 3
f(4) = 3x² - 10
= 3(42) - 10 = 38
\(\therefore\) f(2) - f(4) = 3 - 38 = 35

We have Area of the quadrilateral

=\(\frac { 1 }{ 2 } \) [(-12 - 30 - 28 -10) - (+ 10 + 28 + 30 + 12)]
\(\frac { 1 }{ 2 } \) [-80 - (80)]
\(\frac { 1 }{ 2 } \)[-160] = -80 = 80 square units.
(∵ Area is always +ve).

Given the number 107,
It is a positive odd integer.
Let a = 107 and b = 4
Applying division algorithm we have,
107 = 4q + r where 0 < r < 4
The possible r = 0, 1, 2, 3.
But 107 is odd, the remainders cannot be 0 or 2.
i.e. 4q or 4q + 2 is not possible to express 107.
The other possibilities are 4q + 1 or 4q + 3
Suppose 4q + 1 = 107
4q = 107- 1
= 105
\(q=\frac{106}{4}\) not a natural numbers
Only possibility is 107 = 4q + 3.
107 - 4q + 3
4q = 107 - 3 = 104
\(q=\frac{104}{4}=26\)
q = 26

A = {1, 2),B = {1, 2, 3, 4}, C = {5, 6},D = {5, 6, 7, 8}
A x C = {1, 2} x {5, 6}
\(\mathrm{A} \times \mathrm{C}=\{(1,5),(1,6),(2,5),(2,6)\}\)
B x D = { 1, 2, 3, 4} x { 5, 6, 7, 8}
\(\begin{array}{r} \mathrm{B} \times \mathrm{D}=\{({1,5}),(1,6),(1,7),(1,8) ({2,5}),(2,6),(2,7),(2,8) (3,5),(3,6),(3,7),(3,8) (4,5),(4,6),(4,7),(4,8)\} \end{array}\)
(A x C) is a subset of (B x D)