Answer :
To find the zeros of this function, we must first set the entire function equal to 0
f(x) = x² - 2x - 15 = 0
Since this is a quadratic function, we must use the quadratic formula, which is:
[tex] \frac{-b +/- \sqrt{b^{2} - 4(a)(c) } }{2a} [/tex]
Let's assign a, b, and c using our first function
x² means a = 1 (because it could be written as 1x²)
-2x means b = -2
-15 means c = -15
Now let's plug those in:
[tex] \frac{-(-2) +/- \sqrt{(-2)^{2} - 4(1)(-15) } }{2(1)} [/tex]
which simplifies to:
[tex] \frac{2 +/- \sqrt{(4 + 60} }{2} [/tex]
Simplified further:
[tex] \frac{2 +/- \sqrt{(64} }{2} [/tex]
[tex]\frac{2 +/- 8 }{2} [/tex]
And divide it by the 2 on the bottom gives us:
[tex]2 +/- 4[/tex]
2+4 = 6
2-4 = -2
So the zeros of this function are -2 and 6
f(x) = x² - 2x - 15 = 0
Since this is a quadratic function, we must use the quadratic formula, which is:
[tex] \frac{-b +/- \sqrt{b^{2} - 4(a)(c) } }{2a} [/tex]
Let's assign a, b, and c using our first function
x² means a = 1 (because it could be written as 1x²)
-2x means b = -2
-15 means c = -15
Now let's plug those in:
[tex] \frac{-(-2) +/- \sqrt{(-2)^{2} - 4(1)(-15) } }{2(1)} [/tex]
which simplifies to:
[tex] \frac{2 +/- \sqrt{(4 + 60} }{2} [/tex]
Simplified further:
[tex] \frac{2 +/- \sqrt{(64} }{2} [/tex]
[tex]\frac{2 +/- 8 }{2} [/tex]
And divide it by the 2 on the bottom gives us:
[tex]2 +/- 4[/tex]
2+4 = 6
2-4 = -2
So the zeros of this function are -2 and 6
The zeros of the given function is required.
The zeros of the polynomial are [tex]5,-3[/tex].
The function is
[tex]f(x)=x^2-2x-15[/tex]
[tex]a=1[/tex]
[tex]b=-2[/tex]
[tex]c=-15[/tex]
Using the quadratic formula
[tex]x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\Rightarrow x=\frac{-\left(-2\right)\pm \sqrt{\left(-2\right)^2-4\times1\times \left(-15\right)}}{2\times 1}\\\Rightarrow x=\dfrac{2\pm8}{2}\\\Rightarrow x=5,-3[/tex]
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