The sum of a number and its reciprocal is 10/3, then the numbers are

Let the roots be ? and ? .
Then, ? + ? = 8 ......(i)
? - ? = 4 ..(ii)
On solving Eqs. (i) and (ii), we get
? = 6, ? = 2
? Required equation is
x² - (? + ?) x + (? ?) = 0
? x² - (6 + 2)x + 6 x 2 = 0
? x² - 8x + 12 = 0

We know that, AM ? GM
? ?a + 1/?a ? 2
Here, ?x² - x + 1 + 1/?x² - x + 1 ? 2
? 2 - x² ? 2
? x² ? 0
? x = 0
Hence, the given equation has only one solution.

Since, p and q are the roots of the equation x² + px + q = 0
Then, p + q = - p
and pq = q
Now, pq = q
? p = 1
Putting the value of p in p + q = - p, we get
1 + q = - 1
? q = - 2

Let ? = 2, ? = 1/4
Then, ? + ? = 2 + 1/4 = 9/4
?? = 2.1/4 = 1/2
Equation having the roots ? and ? is
x² - (? + ?)x + ? ? = 0
? x² - 9/4x + 1/2 = 0
? 4x² - 9x + 2 = 0

Let ? and ? be the roots of the quadratic equation 2x² - 11x + 5 = 0
? ? + ? = -(-11)/2 = 11/2 ...(i)
and ?? = 5/2
Now, (? - ?)² = (? + ?)² - 4??
= (11/2)² - 4(5/2) = 121/4 - 20/2
= 121 - 40/4 = 81/4 = (9/2)²
? Difference of roots = (? - ?) = 9/2 = 4.5

? + ? = - p, ? ? = 8
Given, ? - ? = 2
On squaring both sides, we get
(? - ?)² = 4
? (? + ?)² - 4?? = 4
? p² - 32 = 4
? p² = 36 ? p = ± 6

x² + x - 20 = 0
      [by factorisation method]
? x² + 5x - 4x - 20 = 0
? x(x + 5) - 4(x + 5) = 0
(x + 5) (x - 4) = 0
? x = -5 or 4
and y² - y - 30 = 0
? y² - 6y + 5y - 30 = 0
? y(y - 6) + 5(y - 6) = 0
? (y - 6) (y + 5) = 0
? y = 6 or - 5
Hence, y ? x or x ? y

225x² - 4 = 0;
? 225x² = 4 ? x² = 4/225
? x = ?4/225 = ± 2/15, i.e., 2/15 and -2/15
and ?225y + 2 = 0 or ?225y = -2
On squaring both sides, we get
(?225y)² = (-2)²
? 225y = 4
? y = 4/225
So, relation cannot be established because 4/225 lies between 2/15 and -2/15.

x - ?121 = 0
? x = ?121 ? x = ± 11
and y² - 121 = 0 ? y² = 121
? y = ?121 = ± 11
? x = y

x² - 16 = 0
? x² = 16 ? x = ?16 = ± 4
and y² - 9y + 20 = 0
y² - 4y - 5y + 20 = 0
? y(y - 4) - 5(y - 4) = 0
? y = 5, 4
? y ? x or x ? y