Question

If a and b are positive integral numbers and each of the equations x2 + ax + 2b = 0, and x2 + 2bx + a=0 have real roots, then the smallest possible value
of (a + b) is​

Answer 1

Given :  a and b are positive integral numbers and each of the equations x2 + ax + 2b = 0, and x² + 2bx + a=0 have real roots,

To Find : smallest possible value  of (a + b) is​

Solution:

x²  + ax + 2b = 0

Real roots

if  a² - 4(1)(2b)  ≥ 0

=> a² -  8b  ≥ 0

=> a²     ≥  8b

x² + 2bx + a=0

Real roots

if  (2b)² - 4(1)(a)  ≥ 0

=> 4b² - 4a  ≥ 0

=> b² - a  ≥ 0

=>

b²   ≥   a

=> b   ≥   √a

a²     ≥  8b

=> a²     ≥  8√a

Squaring both sides

=> a⁴  ≥  64a

=> a³   ≥  64

=>  a³   ≥  4³

=> a  ≥ 4

b²   ≥   a  

=> b ≥ 2

a  ≥ 4 , b ≥ 2

Smallest possible value of a + b = 4 + 2 = 6

smallest possible value  of (a + b) is​  6

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