Factor trinomials with fractions in binomial squares

[bobisaka]

Unsure of what strategy to use.
x^2 + bx + c does not work with fractions.

please take a look at my working out.. I am only focusing on the left side in this case

 

[pka]

Can you please just state the question that you are asked to work?

 

[bobisaka]

Can you please just state the question that you are asked to work

Solve quadratic equations by completing the square.

2x^2 - 3x = 20

This is my working out for this question.

the thread question is the part I got stuck at, which is when the trinomials becomes a fraction.

 

[Dr.Peterson]

Unsure of what strategy to use.
x^2 + bx + c does not work with fractions.

please take a look at my working out.. I am only focusing on the left side in this case

This is not a factoring problem at all!



What you are doing here is completing the square to solve the equation $x^2-\frac{3}{2}x=10$.

What you do here is to set things up so that you already know what the "factored" form (the perfect square) will be. You don't need to look at $x^2-\frac{3}{2}x+\frac{9}{16}$ and factor it.

Here is how I think of this process: Given $x^2-\frac{3}{2}x=10$, I say, "The first two terms look like the result of squaring a binomial, $(x+h)^2 = x^2+2hx+h^2$. What will h be? Since the second term, $-\frac{3}{2}x$, has to equal $2hx$, our h must be half of $-\frac{3}{2}$, which is $-\frac{3}{4}$. So I want my next line to be $\left(x-\frac{3}{4}\right)^2$.

So the work involved is not to figure out this side by factoring the line above it; it's to add the right term to both sides of the first line so that I will get this on the second.

You're right that the usual method for factoring a trinomial doesn't work well with fractional coefficients. That's why we don't use it.