0 is larger than 0

Algebra Level 2

$\Large{(2x-1)^2-(x+3)^2=0}$

If the sum of all real $\displaystyle x$ can be expressed as $\displaystyle \frac{a}{b}$ , where $a,b$ are positive coprime integers. Find $a + b$ .

1 solution

Adam Phúc Nguyễn
Aug 29, 2015

Method 1

$(2x-1)^2-(x+3)^2=0$

$\Rightarrow (2x-1)^2=(x+3)^2$

$\displaystyle \Rightarrow 2x-1=x+3 \Rightarrow x=-\frac{2}{3}$ or $\displaystyle 2x-1=x-3 \Rightarrow x=4$

Method 2

$(2x-1)^2-(x+3)^2=0$

$\Rightarrow (2x-1-x-3)(2x-1+x+3)=0$

$(x-4)(3x+2)=0$

$\displaystyle\Rightarrow x-4=0 \Rightarrow x=4$ or $\displaystyle 3x+2=0 \Rightarrow x=-\frac{2}{3}$

Method 3

$(2x-1)^2-(x+3)^2=0$

$4x^2-4x+1-x^2-6x-9 = 0$

Then factor, it will be like this:

$(x-4)(3x+2)=0$

Final step: adding

$\large{-\frac{2}{3}+4=\boxed{\frac{10}{3}}}$

Very elegant solution. Keep it up. Upvoted!

Nelson Mandela - 5 years, 9 months ago

Thanks a lot! I'm still waiting for new solutions.

Adam Phúc Nguyễn - 5 years, 9 months ago

I think its the only possible way.

Nelson Mandela - 5 years, 9 months ago

Wow, 3 methods. Impressive :)

Btw, there's some typos on the third line of the first methods, please check carefully and fix :)

Trung Đặng Đoàn Đức - 5 years, 9 months ago