+0

# The equation \$x^2 - (1A)x + A0 = 0\$ has positive integer solutions where \$A\$ is a positive single digit. How many such \$A\$s exist? (Since \$A

0
161
2

(Since A  is representing a digit, if A = 2 then A0 represents the integer 20.)

Jan 28, 2021

#1
+1

x^2 -wx - 10w = 0       (I chnged a to w for clarity)

Discriminant must be > 0

b^2 - 4ac > 0

w^2 -4(1)(-10w) >0

w^2 + 40w >0

w(w + 40) > 0       both w and (w+40)  must be positive for this to be true

w >0   and    w >-40      is restricted to    w > 0        1 2 3 4 5 6 7 8 9    (since it must be positive single digit per Question)

Jan 28, 2021
#2
+120023
+1

Let 1A   =    10  +  A

Let  A0   =    10A

So.....if we  have  integer  solutions, the  discriminant  must  be a  perfect square

So

( 10+A)^2   -  4 (10A)       =

100 + 20A  + A^2  -  40A   =

A^2  - 20A  +  100     =

(A - 10) ( A  -10)   =

(A   - 10)^2

This  will   be a perfect square for any integer    A  >   0

But since A  must  be a single digit....then....

A   can  be any positive  integer from   1  -  9   inclusive

Jan 28, 2021