+160 Given a quadratic equation: x^2 - sqrt(5a^2 - 26a - 8) * x - (a^2 - 4a + 9) = 0
It has 2 integer roots. If a is an integer as well, what is the value of a?
edit: thank you, I really appreciate your help! :)
edit 2: thank you for taking the time to correct it!
+128408 To have two roots.....the discriminant must be a perfect square > 0.....so...
5a^2 - 26a - 8 + 4 (1)(a^2 - 4a + 9) simplify
9a^2 - 42a + 28
Graphing this...it will be > 0 and a perfect square when the integer value for a = 6
So the polynomial is
x^2 - 4x - 21 factor ( x - 7) ( x + 3) and the roots are - 3 and 7
CORRECTED
