Math 3118 Exam 4 Solutions

(20 points) 1. Show that the polynomial p(x)=6x^3 -17x^2+x+10 has exactly three real zeros.
(Possible hint #1: p(-2)=-108, p(-1)=-14, p(0)=10, p(1)=0, p(2)=-8, p(3)=22.)

(Possible hint #2: Factor p(x).)

Proof using hint #1: Since p(-2)=-108, and p(0)=10, the IVT says that there a number c bewteen -2 and 0 such that p(c)=0, it is our first real zero. Our second real zero is x=1, since p(1)=0, and our third is between 2 and 3 by the IVT since p(2)=-8, p(3)=22.

Proof using hint #2: Since x=1 is a zero, we factor,

6x^3 - 17x^2 + x + 10=(x-1)(2x-5)(3x+2), so the three real zeros are 1, 5/2, -2/3.

(20 points) Find rational numbers a and b such that x=a+bi satisfies 5x^2+4=4x, or show that such rational numbers do not exist.

Solution: The quadratic formula gives solutions of x=(4\pm \sqrt(4^2-4*5*4))/10= (4\pm \sqrt(-64))/10=(4\pm 8i)/10, so a=2/5, b=4/5 works.

(20 points) TRUE or FALSE? (Please give your reason in complete sentences.) The product of two infinite non-repeating decimals must be an infinite non-repeating decimal.

Solution: FALSE. \sqrt(2) is an infinite non-repeating decimal. But \sqrt(2)*\sqrt(2)=2 is rational, here a terminating decimal.

(20 points) 4. Explain in your own words what it means when your calculator says that \sqrt{3}=1.7320508. Is it correct?

Solution: No your calculator is not correct when it says \sqrt{3}=1.7320508. We know that \sqrt(3) is irrational, so it could not have a terminating decimal representation. The calculator is giving a rational approximation to \sqrt(3), as a terminating decimal, which agrees with \sqrt(3) to 7 decimal places.

(20 points) 5. Prove that the square root any positive irrational number is an irrational number.

Solution: Let x be a positive irrational number, let's show that \sqrt(x)=p/q is impossible for integers p and q. If \sqrt(x)=p/q, then x=p^2/q^2, so x would be a quotient of integers, which is impossible, since x is irrational.