Respuesta :
[tex]f(x)=x^2-5x-24\\\\x^2-5x-24=0\\\\x^2+3x-8x-24=0\\\\x(x+3)-8(x+3)=0\\\\(x+3)(x-8)=0\iff x+3=0\ \vee\ x-8=0\\\\\boxed{x=-3\ \vee\ x=8}[/tex]
The zeros of any equation lie on the x axis.On x axis y=0.The zeros can be found by substituting y=0 and finding the corresponding y values.
The given equation is :
[tex]y=x^{2}-5x-24[/tex]
Comparing it with quadratic equation [tex]y=ax^{2}+bx+c[/tex]
We have a=1,b=-5,c=-24.
To find the zeros we substitute y=0.
[tex]x^{2}-5x-24=0.[/tex]
WE need to find two numbers which on multiplication will give us ac that is 1(-24)=-24.The same numbers on addition should give us b=-5.
The two numbers are -8 and 3.-8 times 3=-24 and -8+3=-5.
The factors of the equation are (x+8)(x-3)=0
making both factors =0 we have:
x+8=0 and x-3=0
Solving for x
x=-8 and x=3.
The two zeros of the equation are -3,8.