Algebra Know-It-ALL

368 Quadratic Equations with Real Roots

The fact that the above general equation is a quadratic might not appear obvious, but

when we multiply out the left side of the equation using the product of sums rule, the truth

is revealed. Start with the expression

(px+q)(rx+s)

Multiplying this out according to the product of sums rule gives us

pxrx+pxs+qrx+qs

Using the commutative law for multiplication in the first two terms and then simplifying the

first term, we get

prx^2 +psx+qrx+qs

The distributive law “in reverse” allows us to rewrite this as

prx^2 + (ps+qr)x+qs

When we set this equal to 0, we get an equation in polynomial standard form:

prx^2 + (ps+qr)x+qs= 0

Are you confused?

If you have trouble seeing why the above equation is in polynomial standard form, you can rename the

coefficients and constants like this:

pr=a

ps+qr=b

qs=c

By substitution, you get

ax^2 +bx+c= 0

If you wonder about the requirement that p≠ 0 and r≠ 0 in the binomial factor form, the reason for this

restriction should be clear now. If p= 0 or r= 0, then pr= 0, the term containing x^2 vanishes, and you

have the one-variable first-degree equation

(ps+qr)x+qs= 0

What are the roots?

If we find a quadratic equation in binomial factor form and we want to find the roots, we’re

in luck! Here is the binomial factor form again, for a quadratic in the variable x:

(px+q)(rx+s)= 0