x² - 3x - 2 = 0
x² - 3x = 0 (Shifting the Constant to RHS)
x² - 3x + \({ \left( \frac { 3 }{ 2 } \right) }^{ 2 }\) = 2 + \({ \left( \frac { 3 }{ 2 } \right) }^{ 2 }\) (Add [\(\frac {1}{2}\)(co-efficient of x)]² to both sides)
\({ \left( x-\frac { 3 }{ 2 } \right) }^{ 2 }=\frac { 17 }{ 4 } \) (writing the LHS as complete square)
\(x-\frac { 3 }{ 2 } =\pm \frac { \sqrt { 17 } }{ 2 } \) (Taking the square root on both sides)
x = \(\frac { 3 }{ 2 } +\frac { \sqrt { 17 } }{ 2 } \) or x = \(\frac { 3 }{ 2 } -\frac { \sqrt { 17 } }{ 2 } \)
Therefore, x = \(\frac { 3+\sqrt { 17 } }{ 2 } , \frac { 3-\sqrt { 17 } }{ 2 } \)

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

Let p(x) = x² + 18x + 12 = (x + 2)(x + 6)
p(-2) = 4 - 16 + 20 = 0
p(-6) = 36 - 48 + 12 = 0
Therefore -2 and -6 are zero of p(x) = x² + 8x + 12

\(\sqrt { 256{ \left( x-a \right) }^{ 8 }{ \left( x-b \right) }^{ 4 }{ \left( x-c \right) }^{ 16 }{ \left( x-d \right) }^{ 20 } } \) = 16|(x - a)⁴(x - b)² (x - c)⁸ (x - d)¹⁰|

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.

Step 1: Draw the graph of y = 2x² by preparing the table of values as below

Step 2 : To solve 2x² - x - 6 = 0, subtract 2x² - x - 6 = 0 from y = 2x²

The equation y = x + 6 represents a straight line. Draw the graph of y = x + 6 by forming table of values as below

Step 3 : Mark the points of intersection of the curve y = 2x² and the line y = x + 6. That is, (–1.5, 4.5) and (2,8)
Step 4 : The x coordinates of the respective points forms the solution set {–1.5,2} for 2x² - x - 6 = 0

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.

Seven years ago, let Swathi's age be x years .
Seven years ago, let Varun's age was 5x² years.
Swathi's present age = x + 7 years
Varun's present age = (5x² + 7) years
3 years hence, we have
Swathi's age = x + 7 + 3 years
=x + 10 years
Varun's age = 5x² + 7 + 3 years
= 5x² + 10 years
It is given that 3 years hence Swathi's age will
be \(\frac{2}{5}\) of Varun's age.
∴ x+10=\(\frac{2}{5}\)(5x²+10)
⇒ x+10=2x²+4
⇒ 2x²-x-6=0
⇒ 2x(x-2)+3(x-2)=0
⇒(2x+3)(x-2)=0
⇒ x-2=0
⇒ x=2(∵2x+3≠0 as x>0)
Hence Swathi's present age = (2 + 7) years
= 9 years
Varun's present age = (5 x 2² + 7) years
= 27 years

Let the ten's digit of the number be x. It is given that the product of the digits is  12.
Unit's digit \(\frac{12}{x}\)
Number =10x+\(\frac{12}{x}\)
It 36 is added to the number the digits interchange their places.
\(\therefore 10x+\frac { 12 }{ x } +36=10\times \frac { 12 }{ x } +x\)
\(\Rightarrow 10x+\frac { 12 }{ x } +36=\frac { 120 }{ x } +x\)
\(\Rightarrow 9x-\frac { 108 }{ x } +36=0\)
⇒9x² - 108 + 36x = 0
⇒X²+ 4x - 12 = 0
⇒ (x + 6)(x - 2) = 0 (∵ (x + 6) ≠ 0 as x >0)
x=-6,2
But a number can never be (-ve). So, x = 2. The
number is 10x2+\(\frac{12}{2}\)=26

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