Let the length of the side of the chess board be x cm. Then
Area of 64 squares = (x - 4)²
(x - 4)² = 64 x 6.25
⇒ x²-8x+ 16=400
⇒X²- 8x - 384 =0
⇒ X²- 24x + 16x - 384 = 0
⇒(x - 24)(x + 16) = 0
⇒ x=24 cm.

Let the numbers be ∝, β
Sum of the roots = ∝ + β = 15          ...(1)
\(\frac { 1 }{ \alpha } +\frac { 1 }{ \beta } =\frac { 3 }{ 10 } \quad \quad \quad ...(2)\)
\(\\ \frac { +\alpha }{ \alpha \beta } =\frac { 3 }{ 10 } \)
10(∝+ β)= 3∝β    ....(3)
30∝β=10x15=150
Products of the roots =∝β=50 ....(4)
∴ From (1) & (4), we have
x²-15x+50=0
(x-10)(x-5)=0⇒x=10,5

Let the three numbers be x, y, z
From the given data we get the following equations,
3x + y + 2z = 5 .....(1)   
x + 3z - 3y = 2  .....(2)   
2x + 3y - z = 1  .....(3)

Substituting y = 2 in (5), -14 + 7x = 7 gives, z = 3
Substituting y = 2 and z = 3 in (1),  3x + 2 + 6 = 5 we get x = -1
Therefore, x = –1, y = 2, z = 3.

Considering, \(\frac { x }{ 2 } -1=\frac { y }{ 6 } +1\)
\(\frac { x }{ 2 } -\frac { y }{ 6 } \) = 1 + 1  \(\frac { 6x-2y }{ 12 } \) = 2 we get, 3x - y = 12.... (1)
Considering \(\frac { x }{ 2 } -1=\frac { z }{ 7 } +2\)
\(\frac { x }{ 2 } -\frac { z }{ 7 } \) = 1 + 2 gives, \(\frac { 7x-2z }{ 14 } \) = 3 we get, 7x - 2z = 42... (2)
Also, from \(\frac { y }{ 3 } +\frac { z }{ 2 } \) = 13  \(\frac { 2y+3z }{ 6 } \) = 13 we get, 2y + 3z = 78 .(3)
Eliminating z from (2) and (3)

Substituting x = 10 in (1), 30 - y = 12 we get, y = 18
Substituting x = 10 in (2), 70 - 2x = 42 then, z = 14
Therefore, x = 10, y = 18, z = 14.

f(x) = x²+px-4 = 0
If f(-4) = (-4)²+p(-4) = 16-4p-4 = 0
12-4p = 0
-4p = -12
p = 3.
x² + 3x + q = 0 has equal roots, 
∆ = b² - 4ac = 0
3²-4x1xq = 0
9-4q = 0

\(\\ \\ \\ \\ q=\frac { 9 }{ 4 } \)
p = 3, \(\\ \\ \\ \\ q=\frac { 9 }{ 4 } \)

Let f(x) = 2x⁴ + 13x³ + 27x² + 23x + 7,
g(x) = x³ + 3x² + 3x + 1,
h(x) = x² + 2x + 1
which is the least degree polynomial.
Now Dividing f (x) by h(x)

Since the Remainder is zero, h (x) is the GCD of f (x) and h (x)
Now, dividing g (x) by h (x)

Remainder is zero, h (x) is the GCD of g (x) and h (x)
h (x) divides both f(x) and g (x) completely
GCD = x² + 2x + 1

D= b²-4ac
⇒ 4(ac + bd)² - 4(a² + b²)(c² + d²)
⇒ 4[(ac + bd)² - (a² + b²)(c² + d²)]
⇒ 4(a²c² + b²d² + 2acbd - a²c²b²c² - a²d² - b²d²]
⇒ 4[2acbd - a²d² - b²c²]
⇒ 4[a²d² + b²c² - 2adbc]
⇒-4[ ad - bc]²
We have ad≠ bc
∴ ad- be of 0
⇒ (ad - bc)² > 0
⇒ 4(ad - bc)² < 0 ⇒ D < 0
Hence the given equation has no real roots.

9x²-9(a+b)x+(2a²+5ab+2b²)=0
Comparing this with ax² + bx + c = O.
a =9
b = -9(a + b)
c = (2a² + 5ab + 2b²)
∴ ∆=B²-4AC
⇒ 81(a+b)²-36(2a²+5ab+2b²)
⇒ 9a² + 9b² - 18ab
⇒ 9(a - b)2> 0
∴ the roots are real and given by
\(\beta =\frac { -B-\sqrt { D } }{ 2A } =\frac { 9(a+b)+3(a-b) }{ 18 } \)
\(=\frac { 12a+6b }{ 18 } =\frac { 2a+b }{ 3 } \)
\(\beta =\frac { -B-\sqrt { D } }{ 2A } =\frac { 9(a+b)+3(a-b) }{ 18 } \)
\(=\frac { 6a+12b }{ 18 } =\frac { a+2b }{ 3 } \)

x + y + z = 6            ....(1)
2x + 3y + 4z = 20     ...(2)
3x + 2y + 5z = 22   ....(3)

Sub. z = 3 in (5) ⇒ y - 2(3) =-4
y=2
Sub. y = 2, z = 3 in (1), we get
x+2+3=6
x=1
x= 1,y = 2, z = 3

General form of the quadratic equation when the roots are given is
x² - (sum of the roots)x + product of the roots = 0
x² - 9x + 14 = 0

Step 1 : Draw the graph of y = x² + 4x + 3 by preparing the table of values as below

Step 2 : To solve x² + x + 1 = 0, subtract x² + x + 1 = 0 from y = x² + 4x + 3 that is,

The equation represent a straight line. Draw the graph of y = 3x + 2 forming the table of values as below

Step 3 : Observe that the graph of y = 3x + 2 does not intersect or touch the graph of the parabola y = x² + 4x + 3.

Thus x² + x + 1 = 0 has no real roots.