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

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### 1 Answer

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.**