October 13, 2023
October 14, 2023
October 13, 2023
###### Nielit Scientist-B IT 22-07-2017
October 14, 2023
 Question 22

Consider a quadratic equation x2 – 13x + 36 = 0 with coefficients in a base b. The solutions of this equation in the same base b are x = 5 and x = 6. Then b=________.

 A 8 B 9 C 10 D 11
Question 22 Explanation:
x2 – 13x + 36 = 0 ⇾(1)
Generally if a, b are roots.
(x – a)(x – b) = 0
x2 – (a + b)x + ab = 0
Given that x=5, x=6 are roots of (1)
So, a + b = 13
ab=36 (with same base ‘b’)
i.e., (5)b + (6)b = (13)b
Convert them into decimal value
5b = 510
610 = 610
13b = b+3
11 = b+3
b = 8
Now check with ab = 36
5b × 6b = 36b
Convert them into decimals
5b × 6b = (b×3) + 610
30 = b × 3 + 6
24 = b × 3
b = 8
∴ The required base = 8
Question 22 Explanation:
x2 – 13x + 36 = 0 ⇾(1)
Generally if a, b are roots.
(x – a)(x – b) = 0
x2 – (a + b)x + ab = 0
Given that x=5, x=6 are roots of (1)
So, a + b = 13
ab=36 (with same base ‘b’)
i.e., (5)b + (6)b = (13)b
Convert them into decimal value
5b = 510
610 = 610
13b = b+3
11 = b+3
b = 8
Now check with ab = 36
5b × 6b = 36b
Convert them into decimals
5b × 6b = (b×3) + 610
30 = b × 3 + 6
24 = b × 3
b = 8
∴ The required base = 8