Answer:
x=2 or x=8
Step-by-step explanation:
I find it easiest to factor the equation, but you might find it easier to use the quadratic formula. Both will be shown.
<u>Factoring</u>
It is helpful to understand the relationship between the trinomial and its binomial factors:
(x +a)(x +b) = x^2 +(a+b)x +ab
This tells you the constant (ab = 16) will have two factors (a, b) that have a sum equal to the x-coefficient (a+b = -10).
We can list the factorization of 16 to see which factor pairs match that requirement:
16 = -1×-16 = -2×-8 = -4×-4
The sums of these factors are, respectively, -17, -10, -8. The pair with a sum of -10 is the one of interest. This tells us the equation factors as ...
(x -2)(x -8) = 0
The zero product rule tells us that for the product to be zero, one or more of the factors must be zero. The first factor will be zero when x=2; the second factor will be zero for x=8.
The zeros of the quadratic are x=2 or 8.
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<u>Quadratic Formula</u>
The general form of a quadratic equation is often written as ...
![ax^2+bx+c=0](https://tex.z-dn.net/?f=ax%5E2%2Bbx%2Bc%3D0)
And the general solution of this quadratic is given by the formula ...
![x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}](https://tex.z-dn.net/?f=x%3D%5Cdfrac%7B-b%5Cpm%5Csqrt%7Bb%5E2-4ac%7D%7D%7B2a%7D)
Your equation has a=1, b=-10, c=16, so the quadratic formula gives the zeros as ...
![x=\dfrac{-(-10)\pm\sqrt{(-10)^2-4\cdot 1\cdot 16}}{2\cdot 1}=\dfrac{10\pm\sqrt{36}}{2}=5\pm 3](https://tex.z-dn.net/?f=x%3D%5Cdfrac%7B-%28-10%29%5Cpm%5Csqrt%7B%28-10%29%5E2-4%5Ccdot%201%5Ccdot%2016%7D%7D%7B2%5Ccdot%201%7D%3D%5Cdfrac%7B10%5Cpm%5Csqrt%7B36%7D%7D%7B2%7D%3D5%5Cpm%203)
As before, the zeros are 5-3 = 2, and 5+3 = 8.