Respuesta :
The zeros of function are [tex]x = 3 + \sqrt{13} \text{ or } 3 - \sqrt{13}[/tex]
Solution:
Given that we have to find the zeros of given function by completing the square
[tex]y = x^2 - 6x - 4[/tex]
To find the zeros of function, equate to zero
[tex]x^2 - 6x - 4 = 0[/tex]
Completing the square means that we will force a perfect square trinomial on the left side of the equation, then solve for x
Add 4 to both sides of the equation
[tex]x^2 - 6x - 4 + 4 = 0 + 4\\\\x^2 - 6x = 4[/tex]
Divide the coefficient of the x term by 2, then square the result. Add it to both sides of the equation
Here coefficient of x term is -6
[tex](\frac{-6}{2})^2 = 3^2 = 9[/tex]
Add it to both sides of the equation
[tex]x^2 - 6x + 9 = 4 + 9\\\\x^2 - 6x + 9 = 13[/tex]
We now have a perfect square trinomial in the form:
[tex]a^2 -2ab + b^2 = (a - b)^2[/tex]
Here in [tex]x^2 - 6x + 9 = 13[/tex] , a = x and b = 3
[tex]\text{ Substitute } (x-3)^2 \text{ for } x^2 - 6x + 9[/tex]
[tex](x - 3)^2 = 13[/tex]
Take the square root of both sides
[tex]x - 3 = \pm \sqrt{13}[/tex]
Add 3 to both sides of the equation
[tex]x - 3 + 3 = \pm \sqrt{13} + 3\\\\x = 3 \pm \sqrt{13}[/tex]
Therefore,
[tex]x = 3 + \sqrt{13} \text{ or } 3 - \sqrt{13}[/tex]
Thus the zeros of function are [tex]x = 3 + \sqrt{13} \text{ or } 3 - \sqrt{13}[/tex]
Answer:
The answer on top of me was right
Step-by-step explanation:
