Algebra Know-It-ALL

Part Three 525

Adding the quantity (x^2 + 1) to each side, we get

2 x^2 = 2

Dividing this equation through by 2, we obtain

x^2 = 1

It’s apparent, without doing any algebra, that the roots of this are x= 1 or x=−1. Plugging
x= 1 into the first original equation, we get y= 0. Plugging x=−1 into that same equation,
we again get y= 0. This system therefore has two real solutions, (1,0) and (−1,0).

Question 27-4

Consider the following pair of quadratic equations as a two-by-two system:

y=a 1 x^2 +b 1

and

y=a 2 x^2 +b 2

where a 1 and a 2 are nonzero real numbers that are not equal to each other, and b 1 and b 2 are
real numbers. How can we find the real solutions of this system?

Answer 27-4

Once again, let’s call x the independent variable. When we mix the right sides, we get

a 1 x^2 +b 1 =a 2 x^2 +b 2

We can subtract a 2 x^2 from each side and then apply the distributive law to get

(a 1 −a 2 )x^2 +b 1 =b 2

Subtracting b 1 from each side produces

(a 1 −a 2 )x^2 =b 2 −b 1

Dividing through by the quantity (a 1 −a 2 ), which we know is okay because we’ve been told
thata 1 ≠a 2 , we get

x^2 = (b 2 −b 1 ) / (a 1 −a 2 )

This means that

x= [(b 2 −b 1 ) / (a 1 −a 2 )]1/2

or

x=−[(b 2 −b 1 ) / (a 1 −a 2 )]1/2