The value of x in the equation

(	x +	1	)	2	-	3	(	x -	1	)	= 4 is :
		x				2			x

Let, the breadth of the rectangular plot be x m. Then the length of rectangular plot = (x + 8) m
? Area = Length × Breadth = x(x + 8)m² But the area of the plot is given to be 308 m²
? x( x + 8 ) = 308 ? x² + 8x - 308 = 0
? x² + 22x - 14x - 308 = 0
? x( x + 22 ) - 14 ( x - 22 ) = 0
? ( x + 22 ) ( x - 14 ) = 0
? x = 14, -22
But, x = ?22 is not possible, since breath cannot be negative
? x = 14
Hence the breadth of the rectangular plot = 14 m Length of the rectangular plot = (14 + 8) m = 22 m.

According to question , we have
Comparing x² - x + 1 with ax² + bx + c , we get
a = 1, b = ?1, c = 1
Here D = b² - 4ac = ( - 1 )² - 4 (1)(1)
= 1 - 4 = - 3
Since D < 0, so the given expression has no proper linear factor.

Given that : - log₁₀ ( x² - 6x + 45 ) = 2
? x² - 6x + 45 = 10² = 100
? x² - 6x - 55 = 0
? ( x - 11 ) ( x + 5 ) = 0
? x = 11 or x = - 5.
Hence , the values of x are 11 and - 5.

Given equation is
x² - 3x + 1 = 0 ? x² + 1 = 3x
? x² + 1/x = 3
? x²/x + 1/x = 3
? x + 1/x = 3

Given quadratic equation is
7y² - 50y + k = 0
If one root is 7, then it will satisfy the equation i.e putting y = 7 in equation
7 x (7)² - 50 x 7 + k = 0
? 7 x 49 - 350 + k = 0
? 343 - 350 + k = 0
? k = 7

Given equation is
x² + 2(k - 4)x + 2k = 0
On comparing with ax² + bx + c = 0
Here, a = 1, b = 2(k -4), c = 2k
Since, the root are equal, we have D = 0.
b² - 4ac = 0
? 4(k - 4)² - 8k = 0
4(k² + 16 - 8k) - 8k = 0
? 4k² + 64 - 32k - 8k = 0
? 4k² - 40k + 64 = 0
? k² - 10k + 16 = 0
? k² - 8k - 2k + 16 = 0
? k(k - 8) -2 (k - 8) = 0
? (k - 8) (k - 2) = 0
Hence, the value of k 8 or 2.

Given that, the roots of the quadrictic equation are 3 and -1.
Let ? = 3 and ? = -1
Sum of roots = ? + ? = 3 - 1 = 2
Products of roots = ? . ? = (3) (-1) = -3
? Required quadric equation is
x² - (? + ?)x + ? ? = 0
? x² - (2)x + (-3) = 0
? x² - 2x - 3 = 0

Given equation is 4x² - 19x + 12 = 0
Let given equation having the roots 1/? and 1/?,
Then required equuation is
12x² - 19x + 4 = 0