Question

How do I factorise y= -2(x-5)^2 +9 so I get x intercepts

Answer 1

Recall that:

`a^2-b^2=(a+b)(a-b).`

Now, notice that:

`\begin{gathered} -2(x-5)^2+9=-2((x-5)^2-(9)/(2))=-2((x-5)^2-(3^2)/(√(2)^2)) \\ =-2((x-5)^2-((3)/(\sqrt2))^2)=-2((x-5)^2-((3√(2))/(2))^2). \end{gathered}`

Using the first property in the answering tab we get:

`(x-5)^2-((3√(2))/(2))^2=(x-5-(3√(2))/(2))(x-5+(3√(2))/(2))`

Therefore we can rewrite the given equation as follows:

`y=-2(x-5-(3√(2))/(2))(x-5+(3√(2))/(2)).`

Therefore the x-intercepts of the given equation are:

`\begin{gathered} x=-(-5-(3√(2))/(2))=5+(3√(2))/(2) \\ and \\ x=-(-5+(3√(2))/(2))=5-(3√(2))/(2). \end{gathered}`