#1**+1 **

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**+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 | ||

| |||

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