\(x2 + 10x + 25\)
First we can rewrite 25 to 52 and we will check the middle term by multiplying 2ab. This is factoring by the perfect square rule. Now we can check the result of this new problem compared to the original expression.
We now have \(2ab = 2*x*5 = 0\)
Simplify this to \(2ab = 10x =0\)
Now we factor by the perfect square trinomial rule [ \(a2+2ab+b2=(a+b)2a2+2ab+b2=(a+b)2\) where \( a=xa=x \) and \(b=5b=5.\) ] to get: \(( x + 5 ) 2 = 0\).
Next we can set \(x+5\) equal to 0 and solve for \(x \).
Set the factor equal to 0 and we get: \(x+5=0\)
Solve that and you get x = -5
In the question it specifies that we are looking for the probobility that a triangle can exist with side lengths of 1, x and y.
We do know that x and y can be either 0, 1, 2 or 3. We can list that all out:
This gives us a 50/50 chance or 1/2 chance that a triangle with 1,x and y can exists.
Btw I am looking at the problem now and it says the answer is 1/2.
This is just my attempt to explain this but if you see issues with my solving of this problem please point them out.
If you are just here for the answer it is: 1/2