Respuesta :

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[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.


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