SOLUTION: Find all real values of b such that the equation: x^2 + bx + 6b = 0 only has integer roots.

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Question 927517: Find all real values of b such that the equation: x^2 + bx + 6b = 0
only has integer roots.

Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
Find all real values of b such that the equation:
x%5E2+%2B+bx+%2B+6b+=+0 only has integer roots.
Theorem:
Let f%28x%29=ax%5E2+%2B+bx+%2B+c, being a quadratic trinomial with integer coefficients a, b, c. Then, both roots or zeros of f%28x%29 are integers; if, and only if,
(i)
The integer D=b%5E2-4ac is an integer or perfect square.
and
(ii) The leading coefficient a is a divisor of both b and c.

in your case, The leading coefficient a is a divisor of both b and c because a=1
but, we need b; so,
x%5E2+%2B+bx+%2B+6b+=+0+ if discriminant D%3E0 (Positive Discriminant ), we will have Two Real Solutions
D=b%5E2-4ac
b%5E2-4ac%3E0
b%5E2-4%2A1%2A6b%3E0 ........solve for b
b%5E2-24b%3E0+
b%5E2%3E24b
b%3E24
solution is b%3E24
let have first one greater then24:+b=25
highlight%28x%5E2+%2B+25x+%2B+6%2A25+=+0%29 ...only has integer roots
proof:
x%5E2+%2B+25x+%2B+150+=+0
x%5E2+%2B+15x+%2B10x%2B+150+=+0
%28x%5E2+%2B+15x%29+%2B%2810x%2B+150+%29=+0
x%28x+%2B+15%29+%2B10%28x%2B+15+%29=+0
%28x%2B10%29+%28x%2B15%29+=+0
x+=+-15 and x=-10 ->solutions are integer roots

+graph%28+600%2C+600%2C+-20%2C+5%2C+-10%2C+170%2C+x%5E2+%2B+25x+%2B+150%29+