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We want to find the value of $q$ that would make the left-hand side of the equation a perfect square trinomial.
$a_{2}±2ab+b_{2} $
Remember that the first and third terms must be *perfect squares.* Also, if $a_{2}$ and $b_{2}$ are the first and third terms, the middle term must be $twice$ the product of $a$ and $b.$ For the given equation, we will rewrite the first and third terms as perfect squares.
Now, the middle term must be $twice$ the product of $x$ and $5.$
Therefore, if we assume $x =0,$ we have that $q=10.$ Let's rewrite the expression by substituting $10$ for $q,$ and show that it is a perfect square trinomial.
We found that if $q=10,$ the left-hand side of the equation is a perfect square trinomial. What would happen if $q=-10?$ Let's see.
If $q=-10,$ the left-hand side of the equation is also a perfect square trinomial. Thus, there are two possible values for $q$ that make a perfect square trinomial: $q=10$ and $q=-10.$

$x_{2}+qx+25=0$

Substitute$q=10$

$x_{2}+10x+25=0$

FacPosPerfectSquare$a_{2}+2ab+b_{2}=(a+b)_{2}$

$(x+5)_{2}=0$

$x_{2}+qx+25=0$

Substitute$q=-10$

$x_{2}+(-10)x+25=0$

AddNeg$a+(-b)=a−b$

$x_{2}−10x+25=0$

FacNegPerfectSquare$a_{2}−2ab+b_{2}=(a−b)_{2}$

$(x−5)_{2}=0$