x²-4x-5=0
The given equation is quadratic
So we can perform prime factorisation of it.
factors of -5, that add up to give -4 are -5 and 1
therefore replacing -4x by -5x+x
x²-5x+x-5=0
performing factoring by grouping,
(x²-5x)+(x-5)=0
x(x-5)+(x-5)=0
(x+1)(x-5)=0
x+1=0 and x-5=0
Answer is x=-1 and x=5
We must find such numbers, whose sum is equal to -4, and the product is equal to -5.
-4 = 1 + (-5)
-5 = 1 · (-5)
Therefore:
[tex]x^2-4x-5=0\\\\x^2+1x-5x-5=0\\\\x(x+1)-5(x+1)=0\\\\(x+1)(x-5)=0[/tex]
The solutions to this equation are numbers: -1 and 5.
