Part Three 529Now let’s subtract the entire right side of this equation from the left side. When we do that,
we obtain
x^2 + 2 x+ 1 = 0which factors into
(x+ 1)^2 = 0This equation has the single real root x=−1, with multiplicity 2. When we plug that into the
first original equation, we get
y= (x+ 1)^3= (− 1 + 1)^3
= 03
= 0
Therefore, the original system has the single real solution (−1,0), with multiplicity 2.
Question 27-10
Consider the following pair of equations:
y= (x+ 1)^3and
y= (x+ 2)^3How can we find the real solutions of this two-by-two system, if any exist?
Answer 27-10
Let’s multiply both of these equations out to get cubics in polynomial standard form. That
gives us
y=x^3 + 3 x^2 + 3 x+ 1and
y=x^3 + 6 x^2 + 12 x+ 8When we mix the right sides of these equations, we get
x^3 + 3 x^2 + 3 x+ 1 =x^3 + 6 x^2 + 12 x+ 8Subtracting the entire left side from the right side and then switching right-to-left, we obtain
3 x^2 + 9 x+ 7 = 0