Question

(b) Prove that there is a unique integer n for which 2n^2-3n-2=0.

Answer 1

Answer:

The only integer solution is n1=2

Step-by-step explanation:

Let's find the answer by using the following equation applied to quadratic polynomials with the form $an^{2}+bn+c=0$:

$n=\frac{-b\±\sqrt{b^{2}-4ac}}{2a}$

For our case $2n^{2}-3n-2=0$, a=2, b=-3, and c=-2, so:

$n=\frac{-(-3)\±\sqrt{(-3)^{2}-(4*2*(-2))}}{2*2}$

$n=\frac{3\±\sqrt{9+16}}{4}$

$n=\frac{3\±5}{4}$

$n1=\frac{3+5}{4}=2$

$n2=\frac{3-5}{4}=-0.5$

In conclusion, although there are 2 roots (n1, n2), only one of them is an integer, which is n1=2.