A ^ 2x ^ 2 - (3a ^ 2-8a) x + 2A ^ 2-13a + 15 = 0 (where a is a non-zero real number) has at least one integer root to find a I didn't say a was an integer

A ^ 2x ^ 2 - (3a ^ 2-8a) x + 2A ^ 2-13a + 15 = 0 (where a is a non-zero real number) has at least one integer root to find a I didn't say a was an integer

Cross phase multiplication is used
2a^2-13a+15=(a-5)(2a-3)
a^2x^2-(3a^2-8a)x+(a-5)(2a-3)=0
(ax+a-5)(ax+2a-3)=0
Ax + a-5 = 0 or ax + 2a-3 = 0
Then x = (5-a) / A or (3-2a) / A
Because you have to have at least one integer root
Then a = 1

It is known that a & # 178; X & # 178; - (3a & # 178; - 8a) x + 2A & # 178; - 13A + 15 = 0 (where a is a non negative integer) has at least one integer root, and the value of a is obtained

A & # 178; X & # 178; - (3a & # 178; - 8a) x + 2A & # 178; - 13A + 15 = 0 (the following is the cross method) a & # 178; X & # 178; - (3a-8) ax + (2a-3) (a-5) = 0 [ax - (2a-3)]] [ax - (a-5)] = 0, so X1 = (2a-3) / A or x2 = (a-5) / ax1 = (2a-3) / a = 2-3 / a  A is nonnegative integer

Solving equation (2a-1) 2 = 0 (2x-5) 2 + 9 4x2 - (x-1) 2 = 0 by factorization
The next two are squares

Because (2a-1) ^ 2 = 0,
So 2a-1 = 0,
So a = 1 / 2;
（2x-5)^2+9=0,
The equation has no real roots;
Because 4x ^ 2 - (x-1) ^ 2 = 0,
So [(2x) - (x-1)] [2x + (x-1)] = 0,
So (x + 1) (3x-1) = 0,
So x + 1 = 0 or 3x-1 = 0,
So X1 = - 1, X2 = 1 / 3