# For what value of b does 4x^2 - bx + 2 = 0 have integral roots

lfryerda | Certified Educator

From the quadratic formula, we know that the roots of the quadratic equation are {b+-sqrt{b^2-4 cdot 4 cdot 2}}/{2 cdot 4} which can simplify to {b+-\sqrt{b^2-32}}/8.

For these roots to be integral, we have two requirements:

(1) the quantity under the square root needs to be a perfect square and

(2) there needs to be a common factor of 8 in the numerator to cancel the denominator.

Addressing the second requirement first, we let  p=b+-sqrt{b^2-32}.

Now addressing the first requirement, we see that letting b^2-32=m^2 then b^2=m^2+32

where m\in{0,1,2,3,4,5,6,ldots}

which means that b^2\in{32,33,36,41,48,57,68,ldots}

Taking the square root of each value gives us the only potential values for b, which is b^2=36 -> b=+-6 and b^2=81->b=+-9.

Substituting into the original quadratic formula gives no results for b=+-6 but for b=9, then x=2 is an integral root, and for b=-9, then x=-2 is an integral root.