T1W6L2: completing the square

#methods#ae#lesson

completing the square §

i studied jap instead so ill do this later lol :( future thomas here, did it later

recap §

Names	Equation Form
general or y-intercept form	$y=a{x}^2+bx+c$
factorised or x-intercept form	$y=a(x-p)(x-q)$
square or turning point form	$y=a(x-h{)}^2+k$

various features can be quickly identified from each of these forms, and used to sketch a graph.

completing the square §

• completing the square is useful for converting quadratics into turning point form in order to graph or solve them.
• Remember, perfect squares factorise as follows: $x^2+2nx+n^2=(x+n{)}^2$
• 2nx is x coefficient, n^2 is square of half the coefficient, and n is half the coefficient.
• To complete the square, it is faster to use it in the following form: $x^2+2nx=(x+n{)}^2-n^2$
• example
  • consider $2x^2+6x+1$
    1. factorise out the 2 from the first two terms: $2(x^2+3x)+1$
    2. halve the 3, then square it: $3÷2=\frac{3}{2}$, ${\left(\frac{3}{2}\right)}^2=\frac{9}{4}$
    3. form the perfect square $x^2+3x={\left(x+\frac{3}{2}\right)}^2-\frac{9}{4}$
    4. rewrite $2(x^2+3x)+1:2\left({\left(x+\frac{3}{2}\right)}^2-\frac{9}{4}\right)+1$
    5. expand and simplify $2{\left(x+\frac{3}{2}\right)}^2-\frac{7}{2}$

$x^2+2nx=(2+x{)}^2-n^2$ no longer have to explicitly create a perfect square. the new process more simple and streamlined, but if you’re still confused, you can use the old way. solve $2x^2-3x-1=0$ by completing the square $2\left(x^2-\frac{3}{2}x\right)-1$ $x^2-\frac{3}{2}x=(x+\frac{3}{4}{)}^2-\frac{9}{16}$ $=2{\left(x+\frac{3}{4}\right)}^2-\frac{9}{16}+\frac{16}{16}$ $=2{\left(x+\frac{3}{4}\right)}^2+\frac{7}{16}$