Porisma

Worked solution

Complete the square with and without a leading coefficient

Express the following quadratic polynomials in the form a(x+p)2+qa(x+p)^2+q, where a,p,qa,p,q are constants to be found.

(a) x2−5x−45x^2-5x-45

(b) 3x2+4x+63x^2+4x+6

CJC · O Level revision · Example 1

(a) Complete the monic quadratic

Answer

(x−52)2−2054(x-\frac52)^2-\frac{205}4; a=1a=1, p=−52p=-\frac52, q=−2054q=-\frac{205}4.

Group the variable terms

x2−5x−45=(x2−5x)−45\begin{aligned}&x^2-5x-45\\ & =(x^2-5x)-45\end{aligned}

Keep the constant outside the square.

Add and subtract the half-coefficient square

[(x+−52)2−(−52)2]−45=(x−52)2−2054\begin{aligned}&\left[\left(x+\frac{-5}{2}\right)^2-\left(\frac{-5}{2}\right)^2\right]\\ &{}-45\\ & =\left(x-\frac52\right)^2-\frac{205}4\end{aligned}

The printed working halves −5 and subtracts its square to preserve equality.

How did this part go?

(b) Factor out the leading coefficient first

Answer

3(x+23)2+1433(x+\frac23)^2+\frac{14}3; a=3a=3, p=23p=\frac23, q=143q=\frac{14}3.

Make the coefficient inside the bracket one

3x2+4x+6=3(x2+43x)+6\begin{aligned}&3x^2+4x+6\\ & =3\left(x^2+\frac43x\right)+6\end{aligned}

This is the source remark: factorise the variable terms before completing their square.

Preserve the outer factor

3[(x+23)2−(23)2]+6=3(x+23)2−3(23)2+6=3(x+23)2+143\begin{aligned}&3\left[\left(x+\frac23\right)^2-\left(\frac23\right)^2\right]+6\\ & =3\left(x+\frac23\right)^2-3\left(\frac23\right)^2+6\\ & =3\left(x+\frac23\right)^2+\frac{14}3\end{aligned}

The square correction is multiplied by 3 as well.

How did this part go?