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Find the zeros of the function.
Enter the solutions from least to greatest.

f(x)=8x^(2)-800
lesser
x=
greater
x=

Find the zeros of the function. $\newline$ Enter the solutions from least to greatest. $\newline$ $f(x)=8 x^{2}-800$ $\newline$ lesser $x=$ $\newline$ greater $x=$

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Evaluate recursive formulas for sequences

Full solution

Q. Find the zeros of the function.

\newline

Enter the solutions from least to greatest.

\newline

f(x)=8 x^{2}-800

\newline

lesser

x=

\newline

greater

x=

Setting up the equation: To find the zeros of the function $f(x) = 8x^2 - 800$ , we need to set the function equal to zero and solve for $x$ . $0 = 8x^2 - 800$
Simplifying the equation: Next, we simplify the equation by dividing both sides by $8$ to isolate the $x^2$ term. $\newline$ $0/8 = (8x^2 - 800)/8$ $\newline$ $0 = x^2 - 100$
Taking the square root: Now, we solve for $x$ by taking the square root of both sides. Remember that taking the square root of a number yields two solutions: one positive and one negative. $\newline$ $\sqrt{0} = \pm\sqrt{(x^2 - 100)}$ $\newline$ $0 = \pm\sqrt{(x^2)} - \sqrt{100}$ $\newline$ $0 = \pm x - 10$
Solving for x: We now have two equations to solve for the two possible values of x: $x - 10 = 0$ and $-x - 10 = 0$
Solving the first equation: Solving the first equation for $x$ gives us: $x - 10 = 0$ $x = 10$
Solving the second equation: Solving the second equation for $x$ gives us: $\ -x - 10 = 0$ $\ -x = 10$ $\ x = -10$
Listing the zeros: We have found the two zeros of the function $f(x) = 8x^2 - 800$ , which are $x = -10$ and $x = 10$ . We list them in ascending order: $\newline$ lesser $x = -10$ $\newline$ greater $x = 10$

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