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

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

Guest Aug 6, 2017
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#1
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if (x+a)^2=x^2+10x+25 what is the value of a?

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

!

asinus  Aug 6, 2017
#2
+1221
+2

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. $$a=5$$

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. $$a=-2x-5$$

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

$$a_1=5$$

$$a_2=-2x-5$$

TheXSquaredFactor  Aug 6, 2017

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