Algebra Know-It-ALL

Part Three 527

Answer 27-6

The best approach here is to multiply out the right sides of both of the binomial-squared
equations, obtaining the quadratics in polynomial standard form. When we do that, we get

y=x^2 + 2 x+ 1

and

y=x^2 − 2 x+ 1

Mixing the right sides of these quadratics gives us

x^2 + 2 x+ 1 =x^2 − 2 x+ 1

Subtracting the quantity (x^2 + 1) from each side, we obtain

2 x=− 2 x

We can add 2x to each side and then divide through by 4 to get x= 0 as the sole root of this
“mixed quadratic.” When we substitute this value for x into either of the original equations,
we get y= 1. Therefore, the system has the single real solution (0,1).

Question 27-7

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

y= (ax+b)^2

and

y= (ax−b)^2

Where a and b are real numbers, neither of which are equal to 0. How can we find the real
solutions of this system?

Answer 27-7

Again, let’s multiply out the right sides of the binomial-squared equations. When we do that,
we obtain

y=a^2 x^2 + 2 abx+b^2

and

y=a^2 x^2 − 2 abx+b^2

Setting the right sides equal gives us the “mixed quadratic”

a^2 x^2 + 2 abx+b^2 =a^2 x^2 − 2 abx+b^2