if (x+a)^2=x^2+10x+25 what is the value of a?

Guest Aug 6, 2017

if (x+a)^2=x^2+10x+25 what is the value of a?


\((x+a)^2=x^2+10x+25\\ (x+{\color{blue}a})^2=(x+\color{blue}5)^2\\ \color{blue}a=5\)


laugh  !

asinus  Aug 6, 2017

I would argue that there is a second solution to this problem, however. Let me demonstrate why:

\((x+a)^2=x^2+10x+25\) Take the square root of both sides of the equation. Of course, taking the square root of a number results in a positive and a negative answer.
\(x+a=\pm\sqrt{x^2+10x+25}\) Let's split these solutions into 2
\(x+a=\sqrt{x^2+10x+25}\) \(x+a=-\sqrt{x^2+10x+25}\)




Now, let's solve each equation separately:


\(x+a=\sqrt{x^2+10x+25}\) The first step is to factor the expression x^2+10x+25. Now, this trinomial is indeed a perfect-square trinomial. I know this because x^2 and 25 are perfect squares; x and 5. If you multiply the sum of x and 5 by 2, then you get 10x. If this condition is ever true with a trinomial, you have a perfect-square trinomial.
\(x+a=\sqrt{(x+5)^2}\) The square root and the square cancel each other out.
\(x+a=x+5\) Subtract x on both sides.


You will notice that the user above also got this answer of a=5. What about the second solution? Well, you'll see. Solve for a in the following equation \(x+a=-\sqrt{x^2+10x+25}\), the second case:


\(x+a=-\sqrt{x^2+10x+25}\) We have already determined previously that \(\sqrt{x^2+10x+25}\) equals \(x+5\). Let's just plug that in to save a few steps.
\(x+a=-(x+5)\) Distribute the negative sign to both terms.
\(x+a=-x-5\) Subtract x on both sides.


Is this actually a solution, though? If you are ever unsure of whether or not a solution is truly a solution, plug it into the original equation:


\((x+a)^2=x^2+10x+25\) Substitute a for -2x-5
\((x+(-2x-5))^2=x^2+10x+25\) Combine the terms of x and -2x
\((-x-5)^2=x^2+10x+25\) Expand (-x-5)^2 using the rule that \((a+b)^2=a^2+2ab+b^2\)
\(x^2+10x+25=x^2+10x+25\) Both sides of the equation are the same, so it should be clear now that both are equivalent. Therefore, -2x-5 is a solution.


Therefore, there are two solutions




TheXSquaredFactor  Aug 6, 2017

7 Online Users


New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.