Question

9x^2-9(a+b)x+(2a^2+5ab+2b^2)=0 Find x

Answer 1

We have
\[9x^2-9(a+b)x+(2a^2+5ab+2b^2)=0 \]
We know standard quadratic equation
\[ax^2+bx+c=0\]
we know the roots are given by quadratic formula
\[x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\]
here
\[a=9\]
\[b=-9(a+b)\]
and
\[c=2a^2+5ab+2b^2\]
let's find the roots using quadratic formula
\[x=\frac{-9(a+b)\pm \sqrt{(9(a+b))^2-4(9)(2a^2+5ab+2b^2)}}{2\times 9}\]
now let's expand the terms
\[x=\frac{-9(a+b)\pm \sqrt{(81a^2+81b^2+162ab-72a^2-180ab-72b^2)}}{18}\]
we get now
\[x=\frac{-9(a+b)\pm \sqrt{(9a^2+9b^2-18ab)}}{18}\]
let's take 9 out of the square root
we get
\[x=\frac{-9(a+b)\pm 3\sqrt{(a^2+b^2-2ab)}}{18}\]
we know that
\[a^2-2ab+b^2=(a-b)^2\]
we get now
\[x=\frac{-9(a+b)\pm 3(a-b)}{18}\]
so we'll have
\[x=\frac{-9(a+b)+ 3(a-b)}{18} , \frac{-9(a+b)- 3(a-b)}{18}\]
Let's simplify
\[x=\frac{-6a-12b}{18} , \frac{-12a-6b}{18}\]
let's divide numerator and denominator by 6
we get
\[x=\frac{-a-2b}{3} , \frac{-2a-b}{3}\]

Answer 2

you explain step by step.

Answer 3

x=−a−2b | −2a−b
------- | -------
3 | 3