Don't even try taking the square root of both sides.

Algebra Level 1

Solve for $x$ :

$(x+5)^2=(x-5)^2$

The answer is 0.

15 solutions

Victor Loh
Mar 18, 2014

By using the algebraic identities $(a+b)^{2}=a^{2}+2ab+b^{2}$ and $(a-b)^{2}=a^{2}-2ab+b^{2}$ we can expand both sides of the equation to obtain

$x^{2}+(2\times5)x+5^{2}=x^{2}-(2\times5)x+5^{2}$

Cancelling $x^{2}+5^{2}$ from both sides of the equation, we have

$(2\times5)x=-(2\times5)x$

Note that since $2\times5\neq0$ , we can divide both sides of the equation by $2\times5$ to obtain

$x=-x$

Adding $x$ to both sides of the equation, we have

$2x=0 \implies x=\frac{0}{2}=\boxed{0}$

and we are done.

Sorry for the lengthy solution, I just wanted to make it clearer to my readers :)

Well, it's certainly clear. :D

Finn Hulse - 7 years, 2 months ago

You dont need to expand both sides to solve it. You can square root both sides of the equation using the fact that if $x^2=a^2$ , then $x=\pm a$ . So, you should take here not $x+5=x-5$ but $x+5=-(x-5)$ and solve for $x$ accordingly.

P.S. -- I have posted the full solution below.... :)

Prasun Biswas - 7 years, 2 months ago

Lengthy solution is good for me. Thanks for lengthy solution. That's helps me to get properly

Polo Dev - 7 years, 2 months ago

yup..............It works.

Tanzir Hasan Mahim - 7 years, 2 months ago

just deduction

Yasser AL Hadloly - 7 years, 2 months ago

so easy question

iqra yasin - 7 years, 1 month ago

A graphical view:

If y=x^2 is a parabola with its vertex on the origin, then y=(x+5)^2 is the parabola, shifted left by 5 spots, and y=(x-5)^2 is the parabola, shifted right by 5 spots.

Imagine the original parent parabola simultaneously slid left and right, so that the right "branch" of one overlaps/intersects the left branch of the other. The solution asked for is when the parabolas intersect. It remains on the y-axis, climbing higher and higher as the parabolas spread apart, where x=0.

https://graphsketch.com/?eqn1 color=1&eqn1 eqn=%28x-5%29^2&eqn2 color=2&eqn2 eqn=%28x%2B5%29^2&eqn3 color=3&eqn3 eqn=&eqn4 color=4&eqn4 eqn=&eqn5 color=5&eqn5 eqn=&eqn6 color=6&eqn6 eqn=&x min=-10&x max=10&y min=-1&y max=30&x tick=1&y tick=1&x label freq=5&y label freq=5&do grid=0&do grid=1&bold labeled lines=0&bold labeled lines=1&line width=4&image w=850&image_h=525

David Lewis - 5 years ago

Prasun Biswas
Mar 25, 2014

I also have a simple and an easy solution to share. We use the fact that $\sqrt{x^2}=\pm x$ and not just $x$ here. So, we can simply square root the equation as follows ---->

$(x+5)^2=(x-5)^2$

On square rooting both sides ---->

$x+5=\pm(x-5)$

Now, if we take $x+5=x-5$ , we would end up getting an equation like $10=0$ which is not possible and also doesn't give us any value of $x$ . So, we take here $x+5=-(x-5)$ . So --->

$x+5=-(x-5) \implies x+5=-x+5 \implies 2x=0 \implies x=\boxed{0}$

P.S -- I think this might be the simplest and shortest complete solution posted here.... :)

Nice solution. No need to complicate it.😀

Robert Fritz - 7 years, 2 months ago

Me too. :D

Finn Hulse - 7 years, 2 months ago

Or simply state the fact that 5 and -5 squared are equal, so x has got to be 0 for it to work.

César Castro - 3 years, 2 months ago

Siddhartha Srivastava
Mar 18, 2014

$(x+5)^2 = (x-5)^2$

$(x+5)^2 - (x-5)^2 = 0$

$((x+5) + (x-5))((x+5) - (x-5)) = 0$ --------------------Since $a^2 - b^2 = (a-b)(a+b)$

$(2x)(10) = 0$

$20x = 0$

Therefore, $x = \boxed{0}$

Great solution!

Finn Hulse - 7 years, 2 months ago

We can simply open the brackets.

Satvik Golechha - 7 years, 2 months ago

No need for all this !! Simply square root both sides of the original equation as I did in my solution below.

Use the fact that if $x^2=a^2$ , then $x=\pm a$ and not just $a$ to solve this !!

Prasun Biswas - 7 years, 2 months ago

x^2+10x+25 = x^2 - 10x + 25----------10x+25 = -10x+25------ 20x+25 = 25----20x=0

ja ta - 7 years, 2 months ago

Thanks

Kahsay Merkeb - 7 years, 2 months ago

Lokesh Sharma
Mar 18, 2014

Taking square root of both sides:

|x + 5| = |x - 5|

Remember to take modulus around the terms when taking square root. In general $\sqrt { { x }^{ 2 } } = |x|$ .

So yes, the problem can be solved by taking square root (in reference to the title of problem).

I think you made a mistake. It should be $\sqrt{x^2}=\pm x \neq |x|$ .

For, example, let us take $x^2=25$ . Then, we have $x=\pm 5$ and $|x|=5$ .

But, then we can see that $\sqrt{x^2}\neq |x|$ as $|x|$ cannot be $(-5)$ which is one of the values of $\sqrt{x^2}$ taken here. I think you need to edit your solution.

Prasun Biswas - 7 years, 2 months ago

We both are right. It's just a matter of notation. The designation "the square root" is often used to refer to the principal square root. The principal square refers to positive square root (not negative one).

Though, when you see a square root you should treat it as principal square root. I lost some marks due to this in exam once. If you Google the graph of y = sqrt(x), you will see that graph lies only above the y axis. Google it .

For more info, see Wikipedia .

If you have doubt, please reply. I love to talk about such things.

Lokesh Sharma - 7 years, 2 months ago

M. Idrees Ahmed
Mar 20, 2014

x2 +25+10x = x2+25 -10x 20x = 0 x = 0

Sumit Pal
Mar 17, 2014

(x+5)(x+5)=(x-5)(x-5) Then Multiple to each other . Then you Will find Both side is (x+5)2=(x-5)2 (X+5)(x+5)=(x-5)(x-5) X2+5x+5x+25=x2-5x-5x+25 X2+10x=x2-10x 0