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I'm revising quadratics from a textbook which shows me the following forms of the quadratic equation (few of which, I recognise). However, there are some others that I'd never seen. The exercises which follow ask me to express different equations in such forms, but I don't know if they give you specific info as in vertex form or they are just an exercise for practising.

$\alpha x^2+ \beta x + \gamma$ (standard form)
$(x+\alpha)^2 + \beta$
$\alpha (x+\beta)^2 + \gamma$
$\alpha-(x+\beta)^2$
$p-q(x+r)^2$
$(\alpha x+\beta)^2+\gamma$

Other than "where $\alpha, \beta, \gamma, p, q, r$ are constants to be found" , the exercise itself doesn't shed much light on me.

For example: "Express $2x^2-12x+3$ in the form $\alpha (x+\beta)^2 + \gamma$ where $\alpha, \beta, \gamma$ are constants to be found"
$2x^2-12x+3=\alpha (x+\beta)^2 + \gamma$
$2x^2-12x+3=\alpha x^2 - 2\alpha \beta x+ \alpha \beta ^2 + \gamma$
Comparing coefficients of $x^2$ , coefficients of $x$ and the constant gives: $2=\alpha$ (1) , $-12= -2\alpha \beta$ (2) , and $3=\alpha \beta ^2 + \gamma$ (3). By substituting $p=2$ in equation (2) gives $q=3$, and by substituting $p$ and $q$ in equation (3) therefore gives $\gamma=-15$
Hence, $2x^2-12x+3= 2(x-3)^2+15$

My question is as regards this one:

$$13+4x-2x^2$$in the form $p-q(x+r)^2$

I don't know how to approach it and, more importantly, what kind of information can I get out of these... are they just variations of the vertex form?

Thank you very much! (First post here)

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  • $\begingroup$ I suspect that each problem gives you an equation in one of the forms (with particular values for the constants) and asks you to write it in one of the other forms. If you edit the question to show us one example exercise and your attempt to answer it perhaps we can help. $\endgroup$ Commented Oct 21, 2019 at 21:33

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The first one is a simple quadratic in standard form. The rest of them are just variations on the vertex form of a quadratic. Specifically, they are of the form $\alpha (x-\beta)^2 + \gamma,\alpha, \beta,\gamma \in\mathbb{R}$.

To express the given equation in the form $p-q(x+r)^2,$ notice that $q=2$ (just expand the square and use the same coefficient-matching method you demonstrated). By the same method, you should get $r=-1$ and $p=15$. So the equation is $15-2(x-1)^2.$

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Hint. Expand $$ p-q(x+r)^2 = p - q(x^2 + 2rx + r^2) = \cdots $$

Then equate coefficients to those in $$ 13+4x-2x^2 $$ and solve the three simultaneous equations in the unknowns $p,q,r$.

You don't need to know the names of the various forms to do this kind of problem.

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  • $\begingroup$ Thank you! I'm not doing highschool homework. I needed to revise quadratics (among other stuff) because I'm flunking a subject. I was curious about all these forms because, just as with linear equations, I had lots of forms (slope-point, equation from a vector and a point, etc). Perhaps all these variations gave me some info I wasn't aware of. $\endgroup$ Commented Oct 21, 2019 at 22:17

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