Question

(2r+3)²+(r+4)²=10 into the form of ax²+bc+c=0...How do you solve that?

Answer 1

It is given that:

$(2r+3)^2+(r+4)^2=10$

Use the following formula to open the two brackets in the LHS.

$(a+b)^2=a^2+2ab+b^2$

Solve the given question as follows:

$\begin{gathered} (2r)^2+2(2r)(3)+3^2+r^2+2(r)(4)+4^2=10 \\ 4r^2+12r+9+r^2+8r+16=10 \\ 5r^2+20r+25-10=0 \\ 5r^2+20r+15=0 \\ r^2+4r+3=0 \end{gathered}$$\begin{gathered} \text{ Hence the form given by }ax^2+bx+c=0\text{ is obtained which is:} \\ r^2+4r+3=0 \end{gathered}$

Hence a=1,b=4,c=3.