I used to believe that 'completing the square' is in Exp Sec 3 syllabus.suiyuan wrote:HiFrekiWang wrote: 5z^2-30z=5(z^2-6z)=5(z^2-6z+9-9)=5(z^2-6z+9)-45=5(z-3)^2-45

Please explain in more details how you get this.

5z^2-30z=5(z^2-6z)=5(z^2-6z+9-9)=5(z^2-6z+9)-45=5(z-3)^2-45

Thank you

Since you asked, here is the explanation.

You should have learnt that a^2-2ab+b^2=(a-b)^2 in Sec 2.

Compare this formula with z^2-6z. we assume z^2 is the a^2 term and 6z is the 2ab term, then b must be 3. Therefore, when we add a 3^2 (which is equal to 9) at the back, the expression z^2-6z+9 will become a complete square (z-3)^2. However, when we introduce an additional +9, we have to subtract a 9 to keep the whole expression unchanged.