Question

i inserted a picture of the question 6, i can give you the answer to the previous question if it helps

Answer 1

For

`w^2-3w=350`

To be a perfect square trinomial, we need to find a such that

`\begin{gathered} (w-a)^2=w^2-3w+b \\ b\to\text{constant} \end{gathered}`

Thus,

`(w-a)^2=w^2-2aw+a^2`

Then,

`\begin{gathered} \Rightarrow w^2-2aw+a^2=w^2-3w+b \\ \Rightarrow-2a=-3 \\ \Rightarrow a=(3)/(2) \end{gathered}`

Furthermore,

`\begin{gathered} \Rightarrow a^2=b \\ \Rightarrow(9)/(4)=b \\ \end{gathered}`

We need to add 9/4 to both sides of the equation, resulting in

`\begin{gathered} w^2-3w+(9)/(4)=350+(9)/(4) \\ \Rightarrow(w-(3)/(2))^2=(1409)/(4) \end{gathered}`